**1.** Find the dimensional formula of $\epsilon_0$

$\\$ $\epsilon_0 = \frac{Coulomb^2}{Newton m^2} = l^2M^{-1}L^{-3}T^4$ $\\$ $\therefore F = \frac{kq_1q_2}{r^2}$

**2.** A charge of $I^o C$ is placed at the top of the your college
building and another equal charge at the top of your
house. Take the separation between the two charges to
be 2 km. Find the force exerted by the charges on each
other. How many times of your weight is this force ?

$\\$ $q_1 = q_2 = q = 1.0 C distance between = 2 km = 1\times 10^3m$ $\\so, force = \frac{kq_1q_2}{r^2} F = \frac{(9\times 10^9) \times 1 \times 1}{(2\times10^3)^2} = \frac{9\times10^9}{2^2\times 10^6} = 2,25 \times 10^3 N$ $\\$ The weight of body = mg = $40\times 10N = 400N$ $\\$ $So, \frac{wt of body}{force between ch arg es} = (\frac{2.25\times10^3}{4\times 10^2})^{-1} = (5.6)^{-1} = \frac{1}{5.6}$

**3.** At what separation should two equal charges, $I^oC$ each,
be placed so that the force between them equals the
weight of a 50 kg person ?

$\\$ q = 1 C, Let the distance be $x$ $\\$ $F = 50 \times 9.8 = 490$ $\\$ $F = \frac{Kq^2}{x^2} \Rightarrow 490 = \frac{9\times 10^9 \times 1^2}{x^2}$ or $x^2 = \frac{9\times 10^9}{490} = 18.36 \times 10^6$ $\\$ $x= 4.29 \times 10^3 m$

**4.** Two equal charges are placed at a separation of 1.0 m.
What should be the magnitude of the charges so that
the force between them equals the weight of a 50 kg
person ?

$\\$ charges ‘q’ each, AB = 1 m $\\$ wt, of 50 kg person = 50 × g = 50 × 9.8 = 490 N $\\$ $F_c = \frac{kq_1q_2}{r^2}$ $\therefore \frac{kq^2}{r^2} = 490N$ $\\$ $\Rightarrow q^2 = \frac{490\times r^2}{9\times 10^9} = \frac{490\times 1 \times 1}{9\times10^9}$ $\\$ $\Rightarrow \sqrt{54.4 \times 10^{-9}} = 23.323 \times 10^{-5} coulomb = 2.3 \times 10^{-4} coulomb$

**5.** Find the electric force between two protons separated
by a distance of 1 fermi $(1 fermi = 10^{15} m).$ The protons
in a nucleus remain at a separation of this order.

$\\$ Charge on each proton = $a= 1.6 \times 10^{-19}coulomb$ $\\$ Distance between charges = $10 \times 10^{-15} metre = r$ $\\$ Force = $\frac{kq^2}{r^2} = \frac{9\times 10^9 \times 1.6\times1.6\times 10^{-38}}{10^{-15}} = 9\times 2.56 \times 10 = 230.4 Newton$

**6.** Two charges $2 \times 10^{-6} C and 10 \times 10^{-6}6 C$ are placed at
a separation of 10 cm. Where should a third charge be
placed such that it experiences no net force due to these
charges ?

$\\$ $q_1 = 2.0 \times 10^{-6} q_2 = 1.0 \times 10^{-6} r= 10cm = 0.1m$ $\\$ Let the charge be at a distance x from $q_1$ $\\$ $F_1 = \frac{Kq_1q}{x^2} F_2 = \frac{kqq_2}{(0.1-x)^2}$ $\\$ $= \frac{9.9\times 2 \times 10^{-6} \times 10^9 \times q}{x^2}$ $\\$ Now since the net force is zero on the charge q. $\Rightarrow f_1 = f_2$ $\\$ $\Rightarrow \frac{kq_1q}{x^2} = \frac{kqq_2}{(0.1-x)^2}$ $\\$ $2(0.1-x)^2 = x^2 \Rightarrow \sqrt{2}(0.1-x) = x$ $\\$ $\Rightarrow x = \frac{0.1\sqrt{2}}{1+\sqrt{2}} = 0.0586 m = 5.86 cm = 5.9 cm$ From larger charge

**7.** Suppose the second charge in t^ie previous problem is
$- 1 \times 10^{-0} C$. Locate the position where a third charge
will not experience a net force.

$\\$ $q_1 = 2\times 10^{-6}c$ $q_2 = -1\times 10^{-6}c$ $r= 10cm = 10\times 10^{-2} m$ $\\$ Let the third charge be a so, $F_{-AC} = -F_{BC}$ $\\$ $\Rightarrow \frac{kQq_1}{r_1^2} = \frac{-KQq_2}{r_2^2} \Rightarrow \frac{2 \times 10^{-6}}{(10+x)^2} = \frac{1\times10^{-6}}{x^2}$ $\\$ $2x^2 = (10+x)^2 \Rightarrow \sqrt{2}x = 10+x(\sqrt{2}-1) = 10 \Rightarrow x= \frac{-10}{1.414-1} = 24.14cm x$ $\\$ So, distance = 24.14 + 10 = 34.14 cm from larger charge

**8.** Two charged particles are placed at a distance 1 cm
apart. What is the minimum possible magnitude of the
electric force acting on each charge ?

$\\$ Minimum charge of a body is the charge of an electron $\\$ $Wo, q= 1.6 \times 10^{19}C$ $\\$ $x= 1 cm = 1 \times 10^{-2} cm$ $\\$ $So, F= \frac{kq_1q_2}{r^2} = \frac{9\times 10^9 \times 1.6\times 1.6 \times 10^{-19} \times 10^{-19}}{10^{-2} \times 10^{-2}} = 23.01\times 10^{-38+9+2+2} = 23.04\times10^{-25} = 2.3\times 10^{-24}$

**9.** Estimate the number of electrons in 100 g of water. How
much is the total negative charge on these electrons ?

$\\$ No. of electrons of 100 g water = $\frac{10\times100}{18} = 55.5 Nos$ Total charge = 55.5 $\\$ No. of electrons in 18 g of $H_2O = 6.023\times 10^{23} \times 10 = 6.023 \times 10^{24}$ $\\$ No. of electrons in 100 g of $H_2O = \frac{6.023\times 10^{24} \times 100}{18} = 0.334\times 10^{26} = 3.334 \times 10^25$ $\\$ Total charge = $3.34 \times 10^{25} \times 1.6 \times 10^{-19} = 5.34 \times 10^6 C$

**10.** Suppose all the electrons of 100 g water are lumped
together to form a negatively charged particle and all
the nuclei are lumped together to form a positively
charged particle. If these two particles are placed
10 0 cm away from each other, find the force of attraction
between them. Compare it with your weight

$\\$Molecular weight of $H_2O = 2\times 1 \times 16 = 16$ $\\$ No. of electrons present in one molecule of $H_2O = 10$ $\\$ $18 gm of H_2O has 6.023 \times 10^23 molecule$ $\\$ $18 gm of H_2O has 6.023 \times 10^23 \times 10 electrons $ $\\$ $100 gm H_2O has \frac{6.023 \times 10^{24}}{18} \times100 electrons$ $\\$ So number of protons $= \frac{6.023 \times 10^{26}}{18}$ protons (since atom is electrically neutral) $\\$ Charge of protons = $\frac{1.6\times 10^{-19}\times 6.023\times 10^{26}}{18} coulomb = \frac{1.6\times 6.023\times 10^7}{18} coulomb$ $\\$ Charge of electrons = $\frac{1.6\times 6.023 \times 10^7}{18} coulomb$ $\\$ hence Electrical force = $\frac{9\times10^9(\frac{1.6\times6.023 \times10^7}{18}) \times (\frac{1.6\times6.023 \times10^7}{18})}{(10\times10^{-2})^2}$ $\\$ $= \frac{8\times 6.023}{18}\times 1.6 \times 6.023 \times 10^{25} = 2.56 \times 10^{25} Newton$

**11.** Consider a gold nucleus to be a sphere of radius
6-9 fermi in which protons and neutrons are distributed.
Find the force of repulsion between two protons situated
at largest separation. Why do these protons not fly apart
under this repulsion ?

$\\$ Let two protons be at a distance be 13.8 femi $\\$ $F = \frac{9\times10^9 \times 1.6 \times 10^{-38}}{(14.8)^2 \times 10^{30}} = 1.2 N$

**12.** Two insulating small spheres are rubbed against each
other and placed 1 cm apart. If they attract each other
with a force of 0.1N, how many electrons were
transferred from one sphere to the other during
rubbing ?

$\\$ $F= 0.1 N$ $\\$ $r= 1 cm = 10^{-2}$ (As they rubbed with each other. So the charge on each sphere are equal) $\\$ $So, F= \frac{kq_1q_2}{r^2} \Rightarrow 0.1 = \frac{kq^2}{(10^{-2})^2} \Rightarrow q^2 = \frac{0.1 \times 10^{-4}}{9 \times 10^9} \Rightarrow q^2 = \frac{1}{9}\times10^{-14} \Rightarrow q= \frac{1}{3} \times 10^{-7}$ $\\$ $1.6 \times 10^{-19}C$ Carries by 1 electron, 1 c carried by $\frac{1}{1.6\times10^{-19}}$ $\\$ $0.33 \times 10^{-7} C Carries by \frac{1}{1.6\times 10^{-19}} \times 0.33\times10^{-7} = 0.208 \times 10^{12} = 2.08 \times 10^{11}$

**13.** NaCl molecule is bound due to the electric force between
the sodium and the chlorine ions when one electron of
sodium is transferred to chlorine. Taking the separation
between the ions to be $ 2-75 \times 10^8$ cm, find the force of
attraction between them. State the assumptions (if any)
that you have made.

$\\$ $F = \frac{kq_1q_2}{r^2} = \frac{9 \times 10^9 \times 1.6\times1.6\times 10^{-19} \times 10^{-19}}{(2.75 \times 10^{-10})^2} = \frac{23.04\times10^{-29}}{7.56\times10^{-20}} = 3.04 \times 10^{-9}$

**14.** Find the ratio of the electric and gravitational forces
between two protons.

$\\$Given: mass of proton = 1.67 $\times 10^{-27} kg = M_p$ $\\$ $k = 9\times 10^9$ Charge of proton = $1.6 \times 10^{-19} c = C_p$ $\\$ $G= 6.67 \times 10^{-11}$ Let the separation be 'r' $\\$ $Fe = \frac{k(C_p)^2}{r^2}, f_g = \frac{G(M_p)^2}{r^2}$ $\\$ Now, Fe: $Fg = \frac{K(C_p)^2}{r^2} \times \frac{r^2}{G(M_p)^2} = \frac{9\times 10^9 \times (1.6 \times 10^{-19})^2}{6.67\times 10^{-11} \times (1.67\times 10^{-27})^2} = 9\times 2.56 \times 10^{38} = 1.24 \times 10^{38}$

**15.** Suppose an attractive nuclear force acts between two
protons which may be written as F = Ce kr /r.
$\\$(a) Write
down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi 1 and that the
repulsive electric force between the protons is just
balanced by the attractive nuclear force when the
separation is 5 fermi. Find the value of C.

$\\$Expression of electrical force F =$ C \times e^{\frac{-kr}{r^2}}$ $\\$Since $e^{-kr}$ is a pure number. So, dimensional formulae of F $= \frac{dimensional formulae of C}{dimensional formulae of r^2}$ $\\$ Or, $[MLT^2][L^2] = dimensional formulae of C= [ML^3T^2]$ $Unit of C = unit of force \times unit of r2 = Newton \times m^2 = Newton-m^2$ $\\$ Since -kr is a number hence dimensional formulae of $\\$ $k= \frac{1}{dim entional formulae of r} = [L^{-1}]$ Unit of k = $m^{-1}$

**16.** Three equal charges, $2 0 \times 10^{-6} C$ each, are held fixed
at the three corners of an equilateral triangle of side
5 cm. Find the Coulomb force experienced by one of the
charges due to the rest two

$\\$ Three charges are held at three corners of a equilateral triangle. $\\$ Let the charges be A, B and C. It is of length 5 cm or 0.05 m $\\$ Force exerted by B on A = $F_1$ $\\$ force exerted by C on A = $F_2$ $\\$ So, force exerted on A = resultant $F_1 = F_2$ $\\$ $\Rightarrow F = \frac{kq_2}{r^2} = \frac{9\times10^9 \times2 \times2 \times2 \times 10^{-12}}{5\times 5 \times 10^{-4}} = \frac{36}{25} \times 10 = 14.4$ $\\$ Now, force on A = 2 × F cos 30° since it is equilateral $\Delta.$ $\\$ $\Rightarrow Force on A = 2 \times 1.44 \times\sqrt{\frac{3}{2}} = 24.94 N.$

**17.** Four equal charges $2.0 \times 10^{-6} C$ each are fixed at the
four corners of a square of side 5 cm. Find the Coulomb
force experienced by one of the charges due to the rest
three.

$\\$ $q_1 = q_2 = q_3 = q_4 = 2 \times 10^{-6}C$ $\\$ $v = 5 cm = 5 \times 10^{-2} m$ $\\$ so force on $\vec{c} = \vec{F}_{CA} + \vec{F}_{CB} cos45^o + 0$ $\\$ $\frac{k(2\times10^{-6})^2}{(5\times10^{-2})^2} + \frac{k(2\times10^{-6})^2}{(5 \times10^{-2})^2} \frac{1}{2\sqrt{2}} = kq^2 (\frac{1}{25\times10^{-4}} + \frac{1}{50\sqrt{2}\times 10^{-4}})$ $\\$ $= \frac{9\times 10^9 \times 4 \times 10^{-12}}{24\times10^{-4}}(1+\frac{1}{2\sqrt{2}})$ = 1.44 (1.35) = 19.49 Force along \% component = 19.49 $\\$ So, Resultant R = $\sqrt{Fx^2+Fy^2} 19.49\sqrt{2} = 27.56$

**18.** A hydrogen atom contains one proton and one electron.
It may be assumed that the electron revolves in a circle
of radius 0-53 angstrom (1 angstrom =$10^{-10}$ m and is
abbreviated as A) with the proton at the centre. The
hydrogen atom is said to be in the ground state in this
case. Find the magnitude of the electric force between
the proton and the electron of a hydrogen atom in its
ground state

$\\$ R = 0.53 A° = $0.53 \times 10^{–10} m$ $\\$ $F= \frac{kq_1q_2}{r^2} = \frac{9\times 10^9 \times 1.6 \times 1.6 \times 10^-38}{0.53\times 0.53 \times 10^{-10} 10^{-10}} = 82.02 \times 10^{-9} N$

**19.** Find the speed of the electron in the ground state of a
hydrogen atom. The description of ground state is given
in the previous problem.

$\\$ Fe from previous problem No. 18 = $8.2 \times 10^{-8} N Ve =?$ $\\$ Now, $M_e = 9.12 \times 10^{-31} kg r=0.53 \times 10 10^{-10} m$ $\\$ Now, Fe = $\frac{M_ev^2}{r} \Rightarrow v^2 = \frac{Fe \times r}{m_e} = \frac{8.2 \times 10^{-8} \times 0.53 \times 10^{-10}}{9.1 \times 10^{-31}} = 0.4775 \times 10^{13} = 4.775 \times 10^{12} \frac{m^2}{s^2}$ $\\$ $\Rightarrow v = 2.18 \times 10^6 \frac{m}{s}$

**20.** Ten positively charged particles are kept fixed on tne
X-axis at points x = 10 cm, 20 cm, 30 cm, ..., 100 cm.
The first particle has a charge $1.0 \times 10^{-8} C$, the second
$8 \times 10^{-8} C$, the third $27 \times 10^ {-8} C$ and so on. The tenth
particle has a charge $1000 \times 10^{-8} C$. Find the magnitude
of the electric force acting on a 1 C charge placed at the
origin.

$\\$ Electric force feeled by 1 c due to $1 \times 10^{-8}c.$ $\\$ $F_1 = \frac{k \times 1 \times 10^{-8} \times1}{(10\times10^{-2})^2} = k \times 10^{-6}N.$ electric force feeled by 1 c due to $8 \times 10^{-8}c.$ $\\$ $F_2 = \frac{k \times 8 \times 10^{-8} \times1}{(23\times10^{-2})^2} = \frac{k \times 8 \times 10^{-8} \times 10^2}{9} = \frac{28k \times 10^{-6}}{4} = 2k \times 10^{-6} N.$ $\\$ Similarly $F_3 = \frac{k \times 27 \times 10^{-8} \times1}{(23 \times 10^{-2})^2} = 3k \times 10^{-6}N$ $\\$ So, $F= F_1 + F_2 + F_3 + ..... + F_10 = k \times 10^{-6} (1+2+3+....+10)N$ $\\$ $k \times 10^{-6} \times \frac{10 \times11}{2} = 55k \times 10^{-6} = 55 \times 9 \times 10^9 \times10^{-6}N = 4.95 \times10^3 N$

**21.** Two charged particles having charge $2.0 \times 10^{-8} C$ each
are joined by an insulating string of length 1 m find the
system is kept on a smooth horizontal table. Find the
tension in the string.

$\\$ Force exerted = $\frac{kq_1^2}{r^2}$ $\\$ $= \frac{9 \times 10^9 \times 2 \times 2 \times 10^{-16}}{1^2} = 3.6 \times 10^{-6}$ is the force exerted on the string.

**22.** Two identical balls, each having a charge of
$2.00 x 10^{-8} C$ and a mass of 100 g, are suspended from
a common point by two insulating strings each 50 cm
long. The balls are held at a separation 5 0 cm apart and
then released. Find (a) the electric force on one of the
charged balls (b) the components of the resultant force
on it along and perpendicular to the string (c) the
tension in the string (d) the acceleration of one of the
balls. Answers are to be obtained only for the instant
just after the release.

$\\$ $q_1 = q_2 = 2 \times 10^{-7}c$ m= 100g $\\$ $l= 50cm = 5 \times 10^{-2}m$ $d= 5 \times 10^{-2}m$ $\\$ (a) Now Electric force $\\$ F = $K\frac{q^2}{r^2} = \frac{9\times 10^9 \times 4 \times 10^{-14}}{25 \times 10^{-4}}N = 14.4 \times 10^{-2}N = 0.144 N$ $\\$ (b) The components of Resultant force along it is zero, because mg balances $T cos\theta$ and so also. $\\$ F = mg = $T sin \theta$ $\\$ (c) Tension on the string $\\$ $ T sin \theta = F$ $ T cos \theta = mg$ $\\$ $Tan \theta = \frac{F}{mg} = \frac{0.144}{100\times 10^{-3} \times 9.8} = 0.14693$ $\\$ But $T cos \theta = 10^2 \times 10^{-3} \times 10 = 1 N$ $\\$ $\Rightarrow T = \frac{1}{cos \theta} = sec \theta$ $\\$ $\Rightarrow T = \frac{F}{sin \theta},$ $\\$ $Sin \theta = 0.145369; Cos \theta = 0.989378;$

**23.** Two identical pith balls are charged by rubbing against
each other. They are suspended from a horizontal rod
through two strings of length 20 cm each, the separation
between the suspension points being 5 cm. In
equilibrium, the separation between the balls is 3 cm.
Find the mass of each ball and the tension in the strings.
The charge on each ball has a magnitude $2.0 \times 10^{-8} C$.

$\\$ $q=2.0 \times 10^{-8}c n= ? T=? Sin \theta = \frac{1}{20}$ $\\$ Force between the charges $\\$ F = $\frac{Kq_1q_2}{r^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8} \times 2 \times 10^{-8}}{(3 \times 10^{-2})^2} = 4\times 10^{-3}N$ $\\$ $mg sin \theta = F \Rightarrow m = \frac{F}{gsin \theta} = \frac{4\times10^{-3}}{10\times(\frac{1}{20})} = 8 \times 10^{-3} = 8gm$ $\\$ $Cos \theta = \sqrt{1-Sin^2 \theta} = \sqrt{1-\frac{1}{400}} \sqrt{\frac{400-1}{400}} = 0.99 = 1$ $\\$ So, $T = mg cos \theta$ Or T = $8 \times 10^{-3} 10 \times 0.99 = 8 \times 10^{-2}M$

**24.** Two small spheres, each having a mass of 20 g, are
suspended from a common point by two insulating
strings of length 40 cm each. The spheres are identically
charged and the separation between the balls at equilibrium is found to be 4 cm. Find the charge on each
sphere.

$\\$ $T Cos \theta = mg$ ...(1) $\\$ $T Sin \theta = Fe$ ...(2) $\\$ Solving, (2)/(1) we get, $tan \theta = \frac{Fe}{mg} = \frac{kq^2}{r} \times \frac{1}{mg}$ $\\$ $\Rightarrow \frac{2}{\sqrt{1596}} = \frac{9\times 10^9 \times q^2}{(0.04)^2 \times 0.02 \times 9.8}$ $\\$ $\Rightarrow q^2 = \frac{(0.04)^2 \times 0.02\times 9.8\times 2}{9 \times 10^9 \times \sqrt{1596}} = \frac{6.27 \times10^{-4}}{9\times10^9 \times 39.95} = 17\times10^{-16}c^2$ $\\$ $\Rightarrow q = \sqrt{17\times10^{-16}} = 4.123 \times 10^{-8}c$

**25.** Two identical pith balls, each carrying a charge q, are
suspended from a common point by two strings of equal
length I. Find the mass of each ball if the angle between
the strings is $2 \theta$ in equilibrium.

$\\$ Electric Force = $\frac{kq^2}{(l sin Q+ l sin Q)^2} = \frac{kq^2}{4l^2 sin^2}$ So, $T Cos \theta = ms (For equilibrium) T sin \theta = Ef Or tan \theta = \frac{Ef}{mg}$ $\\$ $\Rightarrow mg = Ef cot \theta = \frac{kq62}{4l^2 sin^2 \theta} cot \theta = \frac{q^2 cot \theta0}{l^2 sin^2 \theta 16 \pi E_o}$ $\\$ or $m= \frac{q^2 cot \theta}{16 \pi E_o l^2 Sin^2 \theta g} unit.$

**26.** A particle having a charge of $2.0 \times 1.0 ^{-4} C$ is placed
directly below and at a separation of 10 cm from the bob
of a simple pendulum at rest. The mass of the bob is
100 g. What charge should the bob be given so that the
string becomes loose.

$\\$ Mass of the bob = 100 g = 0.1 kg $\\$ So, Tension in the string = $0.1 \times 9.8 = 0.98N.$ $\\$ For the Tension to be 0, the charge below should repel the first bob. $\\$ $\Rightarrow F= \frac{kq_1q_2}{r^2} T-mg+F = 0 \Rightarrow T = mg-f T= mg$ $\\$ $\Rightarrow 0.98 = \frac{9 \times 10^9 \times 2 \times 10^{-4} q^2}{(0.01)^2} \Rightarrow q_2 = \frac{0.98 \times 1 \times 10^{-2}}{9\times 2 \times 10^5} = 0.054 \times 10^{-9}N$

**27.** Two particles A and B having charges q and 2q
respectively are placed on a smooth table with a
separation d. A third particle C is to be clamped on the
table in such a way that the particles A and B remain
at rest on the table under electrical forces. What should
be the charge on C and where should it be clamped ?

$\\$ Let the change on C = q $\\$ So, net force on c is equal to zero $\\$ So $F_{\vec{AC}}+ F_{\vec{BA}} = 0$ But $F_{AC} = F_{BC} \Rightarrow \frac{kqQ}{x^2} = \frac{k2qQ}{(d-x)^2}$ $\\$ $2x^2 = (d-x)^2 \Rightarrow \sqrt{2} x = d-x$ $\\$ $x= \frac{d}{\sqrt{2}+1} = \frac{d}{(\sqrt{2}+1)} \times \frac{(\sqrt{2} -1)}{(\sqrt{2} -1)} = d(\sqrt{2}-1)$ $\\$ For the charge on rest, $F_{AC} + F_{AB} = 0$ $\\$ $(2.414)^2 \frac{kqQ}{d^2} + \frac{kq(2q)}{d^2} = 0 \Rightarrow \frac{kq}{d^2}[(2.414)^2Q + 2q] = 0$ $\\$ $\Rightarrow 2q = -(2.414)^2 Q$ $\Rightarrow Q = \frac{2}{-(\sqrt{2} + 1)^2}q = -(\frac{2}{3+2\sqrt{2}})q = -(0.343)q = -(6-4\sqrt{2})$

**28.** Two identically charged particles are fastened to the two
ends of a spring of spring constant 100 N/m and natural
length 10 cm. The system rests on a smooth horizontal
table. If the charge on each particle is $2.0 \times 10^{-8} C$, find
the extension in the length of the spring. Assume that
the extension is small as compared to the natural length.
Justify this assmuption after you solve the problem.

$\\$ K= 100 $\frac{N}{m} l = 10cm = 10^{-1}m q = 2.0 \times 10^{-8}c Find l =?$ $\\$ Force between them F $= \frac{kq_1q_2}{r^2} = \frac{9\times10^9 2 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} = 36 \times 10^{-5}N$ $\\$ $So, F = -kx or x = \frac{F}{-K} = \frac{36 \times 10^{-5}}{100} = 36 \times 10^{-7} cm = 3.6 \times 10^{-6}m$

**29.** A particle A having a charge of $2.0 \times 10^{-8} C$ is held fixed
on a horizontal table. A second charged particle of mass
80 g stays in equilibrium.on the table at a distance of
10 cm from the first charge. The coefficient of friction
between the table and this second particle is $\mu$ = 0.2.
Find the range within which the charge of this second
particlc may lie.

$\\$ $q_A = 2 \times 10^{-6} C M_b = 80 g \mu = 0.2$ $\\$ Since B is at equilibrium, So, Fe = $\mu R$ $\\$ $\Rightarrow \frac{Kq_Aq_B}{r^2} = \mu R = \mu m \times g$ $\\$ $\Rightarrow \frac{9\times10^9 \times 2 \times10^{-6}}{0.01} = 0.2 \times0.08 \times9.8$ $\\$ $\Rightarrow q_B = \frac{0.2 \times 0.08 \times 9.8 \times 0.01}{9 \times 10^9 \times 2 \times 10^{-6}} = 8.7 \times 10^{-8}C$ Range = $\pm 8.7 \times 10^{-8}C$

**30.** A particle A having a charge of $2.0 \times 10^{-6} C$ and a mass
of 100 g is placed St the bottom of a smooth inclined
plane of inclination 30°. WTiere should another particle
B, having same charge and mass, be placed on the
incline so that it may remain in equilibrium ?

$\\$ $q_1 = 2 \times 10^{-6} C$ Let the distance be r unit $\\$ $\therefore F_{repulsion} = \frac{kq_1q_2}{r^2} = mg sin \theta$ $\\$ For equilibrium $\frac{kq_1q_2}{r^2} = mg sin \theta$ $\\$ $\Rightarrow \frac{9\times 10^9 \times4 \times 10^{-12}}{r^2} = m \times9.8 \times\frac{1}{2}$ $\\$ $\Rightarrow r^2 = \frac{18 \times 4 \times 10^{-3}}{m \times 9.8} = \frac{72\times10^{-3}}{9.8 \times10^{-1}} = 7.34 \times10^{-2} metre$ $\\$ $\Rightarrow r = 2.70924 \times 10^{-1} meter from the bottom.$

**31.** Two particles A and B, each having a charge Q, are
placed a distance d apart. Where should a particle of
charge q be placed on the perpendicular bisector of AB
so that it experiences maximum force ? What is the
magnitude of this maximum force ?

$\\$ Force on the charge particle ‘q’ at ‘c’ is only the x component of 2 forces. $\\$ $So, F_{on c} = F_{CB} Sin \theta + F_{AC} Sin \theta But \mid \vec{F}_{CB} \mid= \mid \vec{F}_{AC} \mid$ $\\$ $= 2 F_{CB} Sin \theta = 2\frac{KQq}{x^2+(\frac{d}{2})^2} \times \frac{x}{[x^2+\frac{d^2}{4}]^{\frac{1}{2}}} = \frac{2k\theta qx}{(x^2+\frac{d^2}{4})^{\frac{3}{2}}} = \frac{16kQq}{(4x^2+d^2)^{\frac{3}{2}}}x$ $\\$ For maximum force $\frac{dF}{dx} = 0$ $\\$ $\frac{d}{dx}(\frac{16kQqx}{(4x^2+d^2)^{\frac{3}{2}}}) = 0 \Rightarrow K[\frac{(4x^2+d^2)-x[\frac{3}{2}[4x^2+d^2]^\frac{1}{2}8x]}{[4x^2+d^2]^3}] = 0$ $\\$ $\frac{K(4x^2+d^2)^{}\frac{1}{2}[(4x^2+d^2)^3-12x^2]}{(4x^2+d^2)^3}$ $\\$ $16x^4+d^4+8x^2d^2= 12x^2$ $d^4+8x^2d^2 = 0$ $\\$ $\Rightarrow d^2 = 0$ $d^2+8x^2=0 \Rightarrow d^2=8x^2 \Rightarrow d=\frac{d}{2\sqrt{2}}.$

**32.** Two particles A and B, each carrying a charge Q, are
held fixed with a separation d between them. A particle
C having mass m and charge q is kept at the middle
point of the line AB. (a) If it is displaced through a
distance x perpendicular to AB, what would be the
electric force experienced by it. (b) Assuming .t « d,
show that this force is proportional to x. (c) Under what
conditions will the particle C execute simple harmonic
motion if it is released after such a small displacement ?
Find the time period of the oscillations if these
conditions are satisfied.

$\\$(a) Let Q = charge on A & B Separated by distance d.
$\\$ q = charge on c. displaced $\perp$ to –AB
$\\$ $sO, FORCE ON 0= \vec{F}_{AB}+\vec{F}_{BO}$
$\\$ $BUT \vec{F}_{AO} Cos \theta = F_{BO}Cos \theta$
$\\$ So, force on ‘0’ in due to vertical component.
$\\$ $\vec{F} = F_{AO} Sin \theta + F_{BO} Sin \theta$ $\mid F_{AO} \mid = \mid F_{\textbf{}O} \mid$
$\\$ $2\frac{KQq}{(\frac{d}{2^2}+x^2)} Sin\theta$ $F= \frac{2KQq}{(\frac{d}{2})^2 +x^2} Sin\theta$
$\\$ $=\frac{4\times 2 \times kQq}{(d^2+4x^2)} \times \frac{x}{[(\frac{d}{2})^2 + x^2]^{\frac{1}{2}}} = \frac{2kQq}{[(\frac{d}{2})^2 + x^2]^{\frac{3}{2}}} x$
= Electric force $\Rightarrow F \propto x$
$\\$ b) When x << d $F= \frac{2kQq}{[(\frac{d}{2})^2 + x^2]^{\frac{3}{2}}}x$ x<

**33.** Repeat the previous problem if the particle C is
displaced through a distance x along the line AB.

$\\$ $F_{AC} = \frac{KQq}{(l+x)^2} F_{CA} = \frac{KQq}{(l-x)^2}$ $\\$ Net force = $KQq[\frac{1}{(l-x)^2}-\frac{1}{(l+x^2)}]$ $\\$ $= KQq[\frac{(l+x)^2-(l-x)^2}{(l+x)^2(l-x)^2}] = KQq[\frac{4lx}{(l^2-x^2)^2}]$ $\\$ x<<< l = d/2 neglecting x w.r.t. l we get $\\$ net f = $\frac{KQq4lx}{l^4} = \frac{KQq4x}{l^3}$ acceleration = $\frac{4KQqx}{ml^3}$ $\\$ Time period = $2\pi \sqrt{\frac{displacement}{acceleration}} = 2 \pi \sqrt{\frac{ml^3}{4KQq}}$ $\\$ $= \sqrt{\frac{4 \pi^2 ml^3 4\pi \epsilon_0}{4Qq}} = \sqrt{\frac{4 \pi^3 ml^3 \epsilon_0}{Qq}} = \sqrt{4 \pi^3 md^3 \epsilon_0 8Qq} = [\frac{\pi^3 md^3 \epsilon_0}{2Qq}]^{\frac{1}{2}}$

**34.** The electric force experienced by a charge of
$1.0 \times 10^{-6} C$ is $1.5 \times 10^{-3} N$. Find the magnitude of the
electric field at the position of the charge.

$\\$ $F_e= 1.5 \times 10^{-3} N, q=1 \times 10^{-6}C, F_e= q \times E$ $\\$ $\Rightarrow E= \frac{F_e}{q} = \frac{1.5 \times 10^{-3}}{1 \times 10^{-6}}= 1.5 \times 10^3 \frac{N}{C}$

**35.** Two
particles A and
B having
charges of
$2.00 \times 10^{-6}C$ and of$ - 4.00 \times 10^{-6}C$ respectively are
held fixed at a separation of 20 0 cm. Locate the point(s)
on the line AB where (a) the electric field is zero (b) the
electric potential is zero.

$\\$ $q_2 = 2 \times 10^{-6} C, q_1^2 = -4\times 10^{-6}C, r= 20cm = 0.2m$ $\\$ $(E_1 electric field due to q_1 , E_2 = electric field due to q_2)$ $\\$ $\Rightarrow \frac{(r-x)^2}{x^2} = \frac{-q_2}{q_1} \Rightarrow \frac{(r-1)^2}{x} = \frac{-q_2}{q_1} = \frac{4\times 10^{-6}}{2 \times 10^{-6}} = \frac{1}{2}$ $\\$ $\Rightarrow (\frac{r}{x}-1) = \frac{1}{\sqrt{2}} = \frac{1}{1.414} = \Rightarrow \frac{r}{x} = 1.414+1 = 2.414$ $\\$ $\Rightarrow x = \frac{r}{2.414} = \frac{20}{2.414} = 8.282 cm$

**36.** A point charge produces an electric field of magnitude
5 0 N/C at a distance of 40 cm from it. What is the
magnitude of the charge ?

$\\$ $EF = \frac{KQ}{r^2}$ $\\$ $5 \frac{N}{C} = \frac{9 \times 10^9 \times Q}{4^2}$ $\\$ $\Rightarrow \frac{4 \times 20 \times 10^{-2}}{9 \times 10^9} = Q \Rightarrow Q = 8.88 \times 10^{-11}$

**37.** A water particle of mass lO.Omg and having a charge
of $1.50 \times 10^{-6}$C stays suspended in a room. What is the
magnitude of electric field in the room ? What is its
direction ?

$\\$ $m = 10, mg = 10 \times 10^{-3}g \times 10^{-3} kg, q = 1.5 \times 10^{-6}C$ $\\$ But $qE = mg \Rightarrow (1.5 \times 10^{-6} ) E = 10 \times 10^{-6} \times 10$ $\\$ $\Rightarrow E = \frac{10 \times 10^{-4} \times 10}{1.5 \times 10^{-6}} = \frac{100}{1.5} = 66.6 \frac{N}{C}$ $\\$ $\frac{100 \times10^3}{1.5} = \frac{10^{5+1}}{15} = 6.6 \times 10^3$

**38.** Three identical charges, each having a value
$1.0 \times 10^{-8}$ C, are placed at the corners of an equilateral
triangle of side 20 cm. Find the electric field and
potential at the centre of the triangle.

$\\$ $q=1.0 \times 10^{-8}C, l = 20cm$ $\\$ E=? V=? $\\$ Since it forms an equipotential surface. So the electric field at the centre is Zero. $\\$ $r= \frac{2}{3} \sqrt{(2\times 10^{-1})^2-(10^{-1})^2} = \frac{2}{3}\sqrt{4\times 10^{-2}- 10^{-2}}$ $\\$ $= \frac{2}{3}\sqrt{10^{-2} (4-1)} = \frac{2}{3} \times 10^{-2} \times 1.732 = 1.15 \times 10^{-1}$ $V = \frac{3\times 9 \times 10^91 \times 10^{-8}}{1\times 10^{-1}} = 23 \times 10^2 = 2.3 \times 10^3V$

**39.** Positive charge Q is distributed uniformly over a circular
ring of radius R. A particle having a mass m and a
negative charge q, is placed on its axis at a distance x
from the centre. Find the force on the particle. Assuming
x « R, find the time period of oscillation of the particle
if it is released from there.

$\\$We know : Electric field ‘E’ at ‘P’ due to the charged ring $\\$ = $\frac{KQx}{(R^2+x^2)^{\frac{3}{2}}} = \frac{KQx}{R^3}$ $\\$ Force experienced ‘F’ = $Q \times E = \frac{q\times K \times Qx}{R^3}$ $\\$ Now, amplitude = x $\\$ So, T= $2 \pi \sqrt{\frac{x}{\frac{KQqx}{mR^3}}} = 2\pi \sqrt{\frac{mR^3x}{KQqx}} = 2 \pi \sqrt{\frac{4\pi \epsilon_0 mR^3}{Qq}} = \sqrt{\frac{4\pi^2 \times 4\pi \epsilon_0 mR^3}{qQ}}$ $\\$ $\Rightarrow T=[\frac{16\pi^3 \epsilon_0 mR^3}{qQ}]^{\frac{1}{2}}$

**40.** A rod of length L has a total charge Q distributed
uniformly along its length. It is bent in the shape of a
semicircle. Find the magnitude of the electric field at
the centre of curvature of the semicircle.

$\\$ $\lambda = Charge per unit length = \frac{Q}{L}$ $\\$ $dq_1 for a length dl = \lambda \times dl$ $\\$ Electric field at the centre due to charge = $k \times \frac{dq}{r^2}$ $\\$ The horizontal Components of the Electric field balances each other. Only the vertical components remain. $\\$ $\therefore$ Net Electric field along vertical $\\$ $d_E = 2 E cos \theta = \frac{Kdq \times cos \theta}{r^2} = \frac{2kCos \theta}{r^2} \times \lambda \times dl$ $[but d\theta = \frac{dl}{r} = dl = rd\theta]$ $\\$ $\Rightarrow \frac{2k\lambda}{r^2}Cos\theta \times rd \theta = \frac{2k \lambda}{r}Cos \theta \times d\theta$ $\\$ or $E = \int_0^{\frac{\pi}{2}} \frac{2k\lambda}{r}Cos\theta \times d\theta = \int_0^{\frac{\pi}{2}} \frac{2k\lambda}{r}Sin\theta = \frac{2k\lambda l }{r} = \frac{2K\theta}{Lr}$ $\\$ but L = $\pi R \Rightarrow r = \frac{L}{\pi}$ $\\$ So, E = $\frac{2k\theta}{L \times (\frac{L}{\pi})} = \frac{2k\pi \theta}{L^2} = \frac{2}{4\pi \epsilon_0} \times \frac{\pi \theta}{L^2} = \frac{\theta}{2 \epsilon_0L^2}$

**41.** A 10 cm long rod carries a charge of + $50 \mu$C distributed
uniformly along its length. Find the magnitude of the
electric field at a point 10 cm from both the ends of the rod.

$\\$ $G=50 \mu C = 50 \times 10^{-6} C$ $\\$ We have, E = $\frac{2KQ}{r} for a charged cylinder.$ $\\$ $\Rightarrow E = \frac{2\times 9 \times 10^9 \times 50 \times 10^{-6}}{5\sqrt{3}} = \frac{9\times 10^{-5}}{5\sqrt{3}} = 1.03 \times 10^{-5}$

**42.** Consider a uniformly charged ring of radius R. Find the
point on the axis where the electric field is maximum.

$\\$Electric field at any point on the axis at a distance x from the center of the ring is $\\$ $E= \frac{xQ}{4\pi \epsilon_0(R^2+x^2)^{\frac{3}{2}}} = \frac{KxQ}{(R^2+x^2)^{\frac{3}{2}}}$ $\\$ Differentiating with respect to x $\\$ $\frac{dE}{dx} = \frac{KQ(R^2 + x^2)^{\frac{3}{2}} - KxQ(\frac{3}{2})(R^2 + x^2)^{\frac{11}{2}}2x}{(r^2 + x^2)^3}$ $\\$ Since at a distance x, Electric field is maximum. $\\$ $\frac{dE}{dx} = 0 \Rightarrow KQ(R^2+x^2)^{\frac{3}{2}} - Kx^2Q3(R^2+x^2)^{\frac{1}{2}} = 0$ $\\$ $\Rightarrow KQ(R^2 + x^2)^{\frac{3}{2}} = Kx^2Q3(R^2+x^2)^{\frac{1}{2}} \Rightarrow R^2 + x^2 = 3x^2$ $\\$ $\Rightarrow 2x^2 = R^2 \Rightarrow x^2 = \frac{R^2}{2} \Rightarrow x = \frac{R}{\sqrt{2}}$

**43.** A wire is bent in the form of a regular hexagon and a
total charge q is distributed uniformly on it. What is the
electric field at the centre ? You may answer this part
without making any numerical calculations.

$\\$Since it is a regular hexagon. So, it forms an equipotential surface. Hence the charge at each point is equal. Hence the net entire field at the centre is Zero.

**44.** A circular wire-loop of radius a carries a total charge Q
distributed uniformly over its length. A small length DL
of the wire is cut off. Find the electric field at the centre
due to the remaining wire.

$\\$ $\frac{Charge}{Unit length} = \frac{Q}{2\pi a} = \lambda ; Charge od dl = \frac{Qdl}{2\pi a}C$ $\\$ Initially the electric field was ‘0’ at the centre. Since the element ‘dl’ is removed so, net electric field must $\\$ $\frac{K \times q}{a^2}$ Where q = charge of element dl $\\$ $E = \frac{Kq}{a^2} = \frac{1}{4\pi \epsilon_0} \times \frac{Qdl}{2\pi a} \times \frac{1}{a^2} = \frac{Qdl}{8\pi^2 \epsilon_0 a^3}$

**45.** A positive charge q is placed in front of a conducting
solid cube at a distance d from its centre. Find the
electric field at the centre of the cube due to the charges
appearing on its surface.

$\\$ We know, Electric field at a point due to a given charge $\\$ $'E' = \frac{Kq}{r^2}$ Where q = charge, r = Distance between the point and the charge $\\$ So, 'E' = $\frac{1}{4\pi \epsilon_0} \times \frac{q}{d^2} [\therefore r = 'd' here]$

**46.** A pendulum bob of mass 80 mg and carrying a charge
of $2 \times 10^{-6}$C is at rest in a uniform, horizontal electric-
field of 20 kV/m. Find the tension in the thread.

$\\$ $E = 20 \frac{kv}{m} = 20 \times 10^3 \frac{v}{m}, m= 80\times 10^{-5}, c= 20 \times 10^{-5} C$ $\\$ $tan \theta = (\frac{qE}{mg})^(-1) [T Sin \theta = mg, T Cos \theta = qe]$ $\\$ $tan \theta = (\frac{2\times 10^{-8} \times 20\times 10^3}{80\times 10^{-6} \times 10})^{-1} = (\frac{1}{2})^{-1}$ $\\$ $1+ tan^2 \theta = \frac{1}{4} + 1 = \frac{5}{4} [Cos \theta = \frac{1}{\sqrt{5}}, Sin \theta = \frac{2}{\sqrt{5}}]$ $\\$ $T Sin \theta = mg \Rightarrow T \times \frac{2}{\sqrt{5}} = 80 \times 10^{-6}\times 10$ $\\$ $\Rightarrow T = \frac{8 \times 10^{-4} \times \sqrt{5}}{2} = 4\times \sqrt{5} \times 10^{-4} = 8.9 \times 10^{-4}$

**47.** A particle of mass m and charge q is thrown at a speed
u against a uniform electric field E. How much distance
will it travel before coming to momentary rest ?

$\\$Given u = Velocity of projection, $\vec{E}$ = Electric field intensity. $\\$q = Charge; m = mass of particle $\\$We know, Force experienced by a particle with charge ‘q’ in an electric field $\vec{E}$ = qE $\\$ $\therefore$ acceleration produced = $\frac{qE}{m}.$ $\\$ As the particle is projected against the electric field, hence deceleration = $\frac{qE}{m}$ $\\$ So, let the distance covered be ‘s' $\\$ Then $v^2 = u^2 + 2as$ [where a = acceleration, v= final velocity] $\\$ Here 0= $u^2 - 2\times \frac{qE}{m} \times S \Rightarrow S = \frac{u^2m}{2qE} units$

**48.** A particle of mass 1 g and charge $2.5 \times 10^{-4}$C is released
from rest in an electric field of $1.2 \times 10^{-4}$N/C. (a) Find the
electric force and the force of gravity acting on this particle.
Can one of these forces be neglected in comparison with
the other for approximate analysis ? (b) How long will it
take for the particle to travel a distance of 40 cm ? (c) What
will be the speed of the particle after travelling this
distance ? (d) Mow much is the work done by the electric
force on the particle during this period ?

$\\$ $m= 1g = 10^{-3}kg, u=0, q=2.5 \times 10^{-4}C; E=1.2 \times 10^{-4}N/c ; S=40cm = 4\times 10^{-1}m$ $\\$ a) $F= qE = 2.5 \times 10^{-4} \times 1.2 \times 10^4= 3N$ $\\$ So, a= $\frac{F}{m} = \frac{3}{10^{-3}} = 3\times 10^3$ $\\$ $E_q = mg = 10^{-3} \times 9.8 = 9.8 \times 10^{-3}N$ b)S= $\frac{1}{2}at^2$ or t= $\sqrt{\frac{2a}{g}} = \sqrt{\frac{2\times 4 \times 10^{-1}}{3\times 10^3}} = 1.63 \times 10^{-2}sec$ $\\$ $v^2 = u^2+2as = 0+2 \times 3 \times 10^3 \times 4 \times10^{-1} = 24 \times 10^2 \Rightarrow v= \sqrt{24\times 10^2} = 4.9\times 10 = 49 \frac{m}{sec}$ $\\$ work done by the electric force w = $F \rightarrow td = 3\times 4 \times 10^{-1} = 12 \times 10^{-1} = 1.2J$

**49.** A ball of mass 100 g and having a charge of
$4.9 \times 10^{-5}$ C is released from rest in a region where a
horizontal electric field of $2.0 \times 10^{4}$ N/C exists, (a) Find
the resultant force acting on the ball, fa) What will be
the path of the ball ? (c) Where will the ball be at the
end of 2 s ?

$\\$ $m= 100g, q=4.9 \times 10^{-5}, F_g = mg, F=qE$ $\\$ $\vec{E} = 2\times 10^4 \frac{N}{C}$ $\\$ So, the particle moves due to the et resultant R $\\$ $R = \sqrt{F_g^2 + F_e^2} = \sqrt{(0.1\times 9.8)^2 + (4.9\times 10^{-5} \times 2\times 10^4)^2}$ $\\$ $\sqrt{0.9604 + 96.04\times 10^{-2}} = \sqrt{1.9208} = 1.3859N$ $\\$ $tan \theta \frac{F_g}{F_e} = \frac{mg}{qE} =1$ So, $\theta = 45^o$ $\\$ $\therefore$ Hence path is straight along resultant force at an angle 45° with horizontal Disp. Vertical = $(1/2) \times 9.8 \times 2 \times 2 = 19.6 m$ $\\$ Disp. Horizonatal = S= $\frac{1}{2}at^2 = \frac{1}{2}\times \frac{qE}{m}\times t^2 = \frac{1}{2}\times \frac{0.98}{0.1}\times 2 \times2 = 19.6 m$ $\\$ Net. Dispt. = $\sqrt{(19.6)^2 + (19.6)^2} = \sqrt{768.32} = 27.7m$

**50.** The bob of a simple pendulum has a mass of 40 g and
a positive charge of $4.0 \times 10^{-6}$ C. It makes 20 oscillations
in 45 s. A vertical electric field pointing upward and of
magnitude $2.5 \times 10^{4}$ N/C is switched on. How much time
will it now take to complete 20 oscillations ?

$\\$ $m= 40g, q= 4\times 10^{-6} C$ $\\$Time for 20 oscillations = 45 sec. Time for 1 oscillation = $\frac{45}{20}sec$ $\\$ When no electric field is applied, T = $2\pi \sqrt{\frac{l}{g}} \Rightarrow \frac{45}{20} = 2\pi \sqrt{\frac{l}{10}}$ $\\$ $\Rightarrow \frac{l}{10} = (\frac{45}{20})^2 \times \frac{1}{4\pi^2} \Rightarrow l= \frac{(45)^2 \times 10}{(20)^2 \times 4\pi^2} = 1.2836$ $\\$ When electric field is not applied, $\\$ $T=2\pi \sqrt{\frac{l}{g-a}}[a=\frac{qE}{m} = 2.5] = 2\pi \sqrt{\frac{1.2836}{10-2.5}} = 2.598$ $\\$ Time for 1 oscillation = 2.598 $\\$ Time for 20 oscillation = $2.598 \times 20 = 51.96 sec = 52 sec$.

**51.** A block of mass m and having a charge q is placed on
a smooth horizontal table and is connected to a wall
through an unstressed spring of spring constant k as
shown in figure (29-E1). A horizontal electric field E
parallel to the spring is switched on. Find the amplitude
of the resulting SHM of the block.

$\\$ F = qE, F = –Kx $\\$ Where x = amplitude $\\$ qE = – Kx or x = $\frac{-qE}{K}$

**52.** A block of mass m containing a net positive charge q is
placed on a smooth horizontal table which terminates in
a vertical wall as shown in figure (29-E2). The distance
of the block from the wall is <± A horizontal electric field
E towards right is switched on. Assuming elastic
collisions (if any) find the time period of the resulting
oscillatory motion. Is it a simple harmonic motion ?

$\\$ The block does not undergo. SHM since here the acceleration is not proportional to displacement and not always opposite to displacement. When the block is going towards the wall the acceleration is along displacement and when going away from it the displacement is opposite to acceleration. Time taken to go towards the wall is the time taken to goes away from it till velocity is $\\$ $d= ut + (\frac{1}{2})at^2$ $\\$ $\Rightarrow d = \frac{1}{2} \times \frac{qE}{m} \times t^2$ $\\$ $\Rightarrow t^2 = \frac{2dm}{qE} \Rightarrow t = \sqrt{\frac{2dm}{qE}}$ $\\$ $\therefore$ Total time taken for to reach the wall and com back (Time period) $\\$ $2t = 2\sqrt{\frac{2md}{qE}} = \sqrt{\frac{8md}{qE}}$

**53.** A uniform electric field of 10 N/C exists in the vertically
downward direction. Find the increase in the electric
potential as one goes up through a height of 50 cm.

$\\$ $E = 10 \frac{n}{c}, S= 50cm = 0.1m$ $\\$ $E = \frac{dV}{dr} or, V= E\times r = 10\times 0.5 = 5cm$

**54.** 12 J of work has to be done against an existing electric
field to take a charge of 0.01 C from A to B. How much
is the potential difference V B - V A ?

$\\$ Now, $V_B - V_A$ = Potential diff = ? Charge = 0.01 C $\\$ Work done = 12 J Now, Work done = Pot. Diff $\times$ Charge $\\$ $\Rightarrow Pot. Diff = \frac{12}{0.01} = 1200 Volt$

**55.** Two equal charges, $2.0 \times 10^{-7}$ C each, are held fixed at
a separation of 20 cm. A third charge of equal magnitude
is placed midway between the two charges. It is now
moved to a point 20 cm from both the charges. How
much work is done by the electric field during the
process ?

$\\$ When the charge is placed at A, $\\$ $E_1= \frac{Kq_1q_2}{r} + \frac{Kq_3q_4}{r}$ $\\$ $\frac{9\times 10^9 (2\times 10^{-7})^2}{0.1} + \frac{9\times 10^9 (2\times 10^{-7})^2}{0.1}$ $\\$ $\frac{2\times 9 \times 10^9 \times 4 \times 10^{-14}}{0.1} = 72 \times 10^{-4}J$ $\\$ When charge is placed at B, $\\$ $E_2 = \frac{Kq_1q_2}{r} + \frac{Kq_3q_4}{r} = \frac{2\times 9 \times 10^9 \times 4 \times 10^{-14}}{0.2} = 36\times10^{-4}J$ $\\$ Work done =$E_1-E_2 = (72-36) \times 10^{-4} = 36 \times 10^{-4}J = 3.6 \times 10^{-3}J$

**56.** An electric field of 20 N/C exists along the A'-axis in
space. Calculate the potential difference $V_R - V_A$ where
the points A and B are given by,
(a) A = ( 0, 0 ); B - ( 4 m, 2 m )
fa) A - ( 4 m, 2 m ) ; B = ( 6 m , 5 m )
(c) A - ( 0, 0 ); B - ( 6 m, 5 m ).
Do you find any relation between the answers of parts
(a), fa) and (c)

$\\$ (a) A = (0, 0) B = (4, 2) $\\$ $V_B- V_A = E \times d = 20 \times \sqrt{16} 80 V$ $\\$ (b) A(4m, 2m), B = (6m, 5m) $\\$ $\Rightarrow$ $V_B- V_A = E \times d = 20 \times \sqrt{(6-4)^2} = 20\times 2 = 40 V$ $\\$ (c) A(0, 0) B = (6m, 5m) $\\$ $\Rightarrow$ $V_B- V_A = E \times d = 20 \times \sqrt{(6-0)^2} = 20\times 6 = 120 V.$

**57.** Consider the situation of the previous problem. A charge
of - $2.0 \times 10^{-4}$ C is moved from the point A to the point
B. Find the change in electrical potential energy
$U_R - U_A$ for the cases (a), fa) and (c).

$\\$ (a) The Electric field is along x-direction Thus potential difference between (0, 0) and (4, 2) is, $\\$ $\delta V = -E \times \delta x = - 20 \times 40 = -80 V$ $\\$ Potential energy $(U_B – U_A )$ between the points = $\delta V \times q$ $\\$ $= -80 \times (-2) \times 10^{-4} = 160 \times 10^{-4} = 0.016J.$ $\\$ (b) A = (4m, 2m) B = (6m, 5m) $\\$ $\delta V = -E \times \delta x = -20 \times 2 = -40V$ $\\$ Potential energy $(U_B – U_A )$ between the points = $\delta V \times q$ $\\$ =$-40 \times (-2\times 10^{-4}) = 80 \times 10^{-4} = 0.008J$ $\\$ (c) A = (0, 0), B = (6m, 5m) $\\$ $\delta V = -E \times \delta x = -20 \times 6 = -120V$ $\\$ Potential energy $(U_B – U_A )$ between the points A and B $\\$ $\delta V \times q = -120 \times (-2 \times 10^{-4}) = 240 \times 10^{-4} = 0.024J$

**58.** An electric field $\vec{E} - (\vec{i} 20 + \vec{j} 30) $N/C exists in the space.
If the potential at the'origin is taken to be zero, find
the potential at (2 m, 2 m).

$\\$ $E = (\hat{i}20 + \hat{j}30) \frac{N}{C}V = at(2m,2m) r = (2i+2j)$ $\\$So, V= $-\vec{E} \times \vec{r} = -(i20+30j)(2\hat{i} + 2j) = -(2 \times 20 + 2 \times 30) = -100v$

**59.** An electric field $\vec{E} - \vec{i}A$x exists in the space, where
A - 10 $\frac{V}{m^2}$. Take the potential at (10 m, 20 m) to be
zero. Find the potential at the origin.

$\\$ $E= \vec{i} \times Ax = 100 \vec{i}$ $\\$ $\int_v^0dv = -\int E \times dl$ $V= -\int_0^{10} 10x \times dx = -\int_0^{10} \frac{1}{2} \times 10 \times x^2$ $\\$ $0-V = -[\frac{1}{2} \times 1000] = -500 \Rightarrow$ V= 500 volts

**60.** The electric potential existing in spavje is
V(x, y, z) - A(ry + yz + zx). (a) Write the dimensional
formula of A. fa) Find the expression for the electric field,
(c) If A is 10 SI units, find the magnitude of the electric
field at ( 1 m, 1 m, 1 m ).

$\\$ V(x, y, z) = A(xy + yz + zx) $\\$ a) A= $\frac{Volt}{m^2} = \frac{ML^2T^{-2}}{ITL^2} = [MT^{-3}I^{-1}]$ $\\$ b) E= $-\frac{\delta V \hat{i}}{\delta x} -\frac{\delta V \hat{j}}{\delta y} -\frac{\delta V \hat{k}}{\delta z} = -[\frac{\delta}{\delta x}A(xy+yz+zx) + \frac{\delta}{\delta y} A(xy+yz+zx) + \frac{\delta}{\delta z} A(xy+yz+zx)]$ $\\$ = $-[(Ay+Az)\hat{i} + (Ax+Az)\hat{j} + (Ay+Ax)\hat{k}] = -A(y+z)\hat{i} + A(x+z)\hat{j} + A(y+x)\hat{k}$ $\\$ (c) A = 10 SI unit, r = (1m, 1m, 1m) $\\$ E = $–10(2)\hat{i} – 10(2) \hat{j} – 10(2) \hat{k} = – 20 \hat{i} – 20 \hat{j} – 20 \hat{k} = \sqrt{20^2 + 20^2 + 20^2} = \sqrt{1200} = 34.64 = 35 \frac{N}{C}$

**61.** Two charged particles, having equal charges of
$2 0 \times 10{-5}$ C each, are brought from infinity to within a
separation of 10 cm. Find the increase in the electric
potential energy during the process.

$\\$ q 1 = q 2 = $2 \times 10^{-5}C$ $\\$ Each are brought from infinity to 10 cm a part d = $10 \times 10^{-2}m.$ $\\$ So work done = negative of work done. (Potential E). $\\$ P.E = $\int_\infty^{10}F \times ds$ P.E = $K \times \frac{q_1q_2}{r} = \frac{9\times 10^9 \times 4 \times 10^{-10}}{10\times 10^{-2}} = 36J$

**62.** Some equipotential surfaces are shown in figure (29-E3).
What can you say about the magnitude and the direction
of the electric field ?

$\\$ (a) The angle between potential E dl = dv $\\$ Change in potential = 10 V = dV $\\$ As E = $\perp$r dV (As potential surface) $\\$ So, $E dl = dV \Rightarrow E dl Cos(90^o+30^o) = -dv$ $\\$ $\Rightarrow E(10\times 10^{-2})cos 120^o = -dV$ $\\$ $\Rightarrow = \frac{-dV}{10 \times 10^{-2} Cos 120^o} = -\frac{10}{10^{-1} \times (\frac{-1}{2})}$ = 200 V/m making an angle 120° with y-axis. $\\$ (b) As Electric field intensity is $\perp$r to Potential surface. $\\$ So, E= $\frac{kq}{r^2}r = \frac{kq}{r} \Rightarrow \frac{kq}{r} = 60v q = \frac{6}{K}$ $\\$ So, E=$\frac{kq}{r^2} = \frac{6\times k}{k\times kr^2}v.m = \frac{6}{r^2}v.m$

**63.** Consider a circular ring of radius r, uniformly charged
with linear charge density $\lambda$. Find the electric potential
at a point on the axis at a distance x from the centre of
the ring. Using this expression for the potential, find the
electric field at this point.

$\\$ Radius = r So, $2\pi r$ = Circumference $\\$ Charge density = $\lambda$ Total charge = $2\pi r \times \lambda $ $\\$Electric Potential = $\frac{Kq}{r} = \frac{1}{4\pi \epsilon_0} \times \frac{2\pi r \lambda}{(x^2 + r^2)^{\frac{1}{2}}} = \frac{r\lambda}{2 \epsilon_0 (x^2+r^2)^\frac{1}{2}}$ $\\$So, Electric field = $\frac{V}{r}Cos \theta$ $\\$ $\frac{r \lambda}{2\epsilon_0(x^2+r^2)^{\frac{1}{2}}} \times \frac{1}{(x^2+r^2)^{\frac{1}{2}}}$ $\\$ $= \frac{r\lambda}{2\epsilon_0(x^2+r^2)^{\frac{1}{2}}} \times \frac{x}{(x^2+r^2)^{\frac{1}{2}}} = \frac{r\lambda x}{2\epsilon_0 (x^2+r^2)^{\frac{3}{2}}}$

**64.** An electric field of magnitude 1000 N/C is produced
between two parallel plates having a separation of 2 0 cm
as shown in figure (29-E4). (a) What is the potential
difference between the plates ? fa) With what minimum
speed should an electron be projected from the lower plate
in the direction of the field so that it may reach the upper
plate ? (c) Suppose the electron is projected from the lower
plate with the speed calculated in part fa). The direction
of projection makes an angle of 60° with the field. Find
the maximum height reached by the electron.

$\\$ $\vec{E} = 1000 \frac{N}{C}$ $\\$ (a) V = E × dl = 1000 $\times \frac{2}{100} = 20V$ $\\$ (b) u = ? $\vec{E} = 1000, = \frac{2}{100}m$ $\\$ a = $\frac{F}{m} = \frac{q\times E}{m} = \frac{1.6 \times 10^{-19} \times 1000}{9.1\times 10^{-31}} = 1.75 \times 10^{14} \frac{m}{s^2}$ $\\$ 0 = $u^2 - 2 \times 1.75 \times 10^{14} \times 0.02 \Rightarrow u^2 = 0.04\times 1.75 \times 10^{14} \Rightarrow u = 2.64 \times 10^6 \frac{m}{s}.$ $\\$ (c) Now, U = u Cos 60° V = 0, s = ? $\\$ $a= 1.75 \times 10^{14} \frac{ m}{s^2} V^2 = u^2-2as$ $\\$ $\Rightarrow s= \frac{(ucos 60^o)^2}{2\times a} = \frac{(2.64 \times 10^6 \times \frac{1}{2})^2}{2\times 1.75 \times 10^{14}} = \frac{ 1.75 \times 10^{12}}{3.5 \times 10^{14}} = 0.497 \times 10^{-2} = 0.005 m = 0.50cm$

**65.** A vmiform field of 2.0 N/C exists in space in .v-direction.
(a).Taking the potential at the origin to be zero, write
an expression for the potential at a general point
(x,y, z).J?o) At which points, the potential is 25 V ?<£c)Tf
the potential at the origin is taken to be 100 V, what
will be Uie expression for the potential at a general
point ? (a) What will be the potential at the origin if the
potential at infinity is taken to be zero ? Is it practical
to choose the potential at infinity to be zero ?

$\\$ E = 2 N/C in x-direction $\\$ (a) Potential aat the origin is O. dV = $- E_x dx - E_y dy - E_z dz$ $\\$ $\Rightarrow v-0 = -2x \Rightarrow x = -2x$ $\\$ (b) (25 – 0) = – 2x $\Rightarrow$ x = – 12.5 m. $\\$ (c) If potential at origin is 100 v, v – 100 = – 2x $\Rightarrow V = -2x + 100 = 100 -2x$ $\\$ (d) Potential at $\infty$ IS 0, $V-V' = -2x \Rightarrow V' - V+2x = 0 + 2\infty \Rightarrow V' = \infty$ $\\$ Potential at origin is $\infty$ No, it is not practical to take potential at $\infty$to be zero.

**66.** How much work has to be done in assembling three
charged particles at the vertices of an equilateral
triangle as shown in figure (29-E5) ?

$\\$ Amount of work done is assembling the charges is equal to the net potential energy. $So, P.E. = U_{12} + U_{13} +U_{23}$ $\\$ = $\frac{Kq_1q_2}{r_{12}} + \frac{Kq_1q_3}{r_{13}} + \frac{Kq_2q_3}{r_{23}}$ = $\frac{K \times 10^{-10}}{r} [4 \times 2 + 4 \times 3+ 3 \times 2]$ $\\$ $\frac{9 \times 10^9 \times 10^{-10}}{10^{-1}} (8+12+6) = 9\times 26 = 234 j$

**67.** The kinetic energy of a charged particle decreases by
10 J as it moves from a point at potential 100 V to a
point at potential 200 V. Find the charge on the particle.

$\\$ K.C. decreases by 10 J. Potential = 100 v to 200 v. $\\$ So, change in K.E = amount of work done $\\$ $\Rightarrow 10J = (200-100)v \times q_0 \Rightarrow 100 q_0 = 10v$ $\\$ $\Rightarrow q_0 = \frac{10}{100} = 0.1C$

**68.** Two identical particles, each having a charge of
$2.0 \times 10^{-4}$ C and mass of 10 g, are kept at a separation
of 10 cm and then released. What would be the speeds
of the particles when the separation becomes large ?

$\\$ m= 10g ; F = $\frac{KQ}{r} = \frac{9 \times 10^9 \times 2 \times 10^{-4}}{10 \times 10^{-2}} f= 1.8 \times 10^{-7}$ $\\$ $F= m \times a \Rightarrow a = \frac{1.8 \times 10^{-7}}{10\times 10^{-3}} = 1.8 \times 10^{-3} \frac{m}{s^2}$ $\\$ $V^2 - u^2 = 2as \Rightarrow V^2 = u^2 + 2as$ $\\$ $V= \sqrt{0+2\times 1.8 \times 10^{-3} \times 10 \times 10^{-2}} = \sqrt{3.6\times 10^{-4}} = 0.6 \times 10^{-2} = 6 \times 10^{-3} \frac{m}{s}.$

**69.** Two particles have equal masses of 5 0 g each and
opposite charges of + $4.0 \times 10^{-8}$ C and - $4.0 \times 10^{-5}$C.
They are released from rest with a separation of TO m
between them. Find the speeds of the particles when the
separation is reduced to 50 cm.

$\\$ $q_1 = q_2 = 4 \times 10^{-5} ; s = 1m, m = 5 g = 0.005 kg$ $\\$ $F= K\frac{q^2}{r^2} = \frac{9 \times 10^9 \times (4 \times 10^{-5})^2}{1^2} = 14.4N.$ $\\$ Acceleration 'a' = $\frac{F}{m} = \frac{14.4}{0.005} = 2880 \frac{m}{s^2}$ $\\$ Now u=0, $s = 50cm= 0.5m,$ $a = 2880 \frac{m}{s^2}, V=?$ $\\$ $V^2 = u^2 + 2as \Rightarrow V^2 = = 2 \times 2880 \times 0.5$ $\\$ $\Rightarrow V= \sqrt{2880} = 53.66 \frac{m}{s} = 54 \frac{m}{s} for each particle$

**70.** A sample of HC1 gas is placed in an electric field of
$2.5 \times 10^4$ N/C. The dipole moment of each HC1 molecule
is $3.4 \times 10^{-30}$ C-m. Find the maximum torque that can
act on a molecule.

$\\$ E = 2.5 × 104 $P = 3.4 \times 10^{-30} \tau = PE sin \theta$ $\\$ $P \times E \times 1 = 3.4 \times 10^{-30} \times 2.5 \times 10^4 = 8.5 \times 10^{-26}$

**71.** Two particles A and B, having opposite charges
$2.0 \times 10^{-6}$ C and $- 2.0 \times 1.0^{-6}$C , are placed at a
separation of TO cm. (a) Write down the electric dipole
moment of this pair, (b) Calculate the electric field at a
point on the axis of the dipole TO cm away from the
centre, (c) Calculate the electric field at a point on the
perpendicular bisector of the dipole and TO m away from -
the centre.

$\\$ (a) Dipolemoment = q × l $\\$ (Where q = magnitude of charge l = Separation between the charges) $\\$ $= 2 \times 10^{-6} \times 10^{-2} cm = 2 \times 10^{-8}cm$ $\\$ (b) We know, Electric field at an axial point of the dipole $\\$ $= \frac{2kp}{r^3} = \frac{2\times 9 \times 10^9 2 \times 10^{-8}}{(1\times 10^{-2})^3} = 36\times 10^7 \frac{N}{C}$ $\\$ (c) We know, Electric field at a point on the perpendicular bisector about 1m away from centre of dipole. $\\$ $= \frac{KP}{r^3} = \frac{9 \times 10^9 2 \times 10^{-8}}{1^3} = 180 \frac{N}{C}$

**72.** Three charges are arranged on the vertices of an
equilateral triangle as shown in figure (29-E6). Find the
dipole moment of the combination.

$\\$Q:72 $\\$ Let –q & –q are placed at A & C $\\$ Where 2q on B, So length of A = d $\\$ So the dipole moment = $(q \times d) = P$ $\\$ So, Resultant dipole moment $\\$ $P= [(qd^2) + (qd)^2 + 2qd \times qd Cos 60^o]^{\frac{1}{2}} = [3q^2d^2]^{\frac{1}{2}} = \sqrt{3}qd = \sqrt{3}P$

**73.** Find the magnitude of the electric field at the point P
in the configuration shown in figure (29-E7) for d » a.
Take 2qa = p

$\\$(a) P = 2qa $\\$ $(b) E_1 Sin \theta = E_2 sin \theta$ Electric field intensity $\\$ $= E_1 Cos \theta + E_2 cos \theta = 2 E_1 Cos \theta$ $\\$ $E_1 = \frac{Kqp}{a^2+d^2} s0, E=\frac{2KPQ}{a^2 + d^2} \frac{a}{(a^2+d^2)^{\frac{1}{2}}} = \frac{2Kq \times a}{(a^2+d^2)^{\frac{3}{2}}}$ $\\$ When a << d $= \frac{2Kqa}{(d^2)^{\frac{3}{2}}} = \frac{PK}{d^3} = \frac{1}{4\pi \epsilon_0} \frac{P}{d^3}$

**74.** Two particles, carrying charges - q and + q and having
equal masses m each, are fixed at the ends of a light
rod of length a to form a dipole. The rod is clamped at
an end and is placed in a uniform electric field E with
the axis of the dipole along the electric field. The rod is .•
slightly tilted and then released. Neglecting gravity find'
the time period of small oscillations,

$\\$ Consider the rod to be a simple pendulum. $\\$ For simple pendulum $T= 2\pi \sqrt{\frac{l}{g}}$ (l = length, q = acceleration) $\\$ Now, force experienced by the charges $\\$ $F= Eq$ Now, acceleration = $\frac{F}{m} = \frac{Eq}{m}$ $\\$ Hence length = a so, Time period = $2\pi \sqrt{\frac{a}{(\frac{Eq}{m})}} = 2\pi \sqrt{\frac{ma}{Eq}}$

**75.** Assume that each atom in a copper wire contributes one
free electron. Estimate the number of free electrons in
a copper wire having a mass of 6-4 g (take the atomic .
weight of copper to be 64 g/mol).

$\\$ 64 grams of copper have 1 mole, 6.4 grams of copper have 0.1 mole $\\$ 1 mole = No atoms, 0.1 mole = (no × 0.1) atoms $\\$ $6 \times 10^{23} \times 0.1 atoms = 6 \times 10^{22} atoms$ $\\$ 1 atom contributes 1 electron $6 \times 10^{22} atoms contributes 6 \times 10^{22} electrons.$