Electric Charges and Fields

Physics

NCERT

1   What is the force between two small charged spheres having charges of $2 *10^{-7} C$ and $3* 10^{-7 }C$ placed $30 cm$ apart in air?

Solution :

Repulsive force of magnitude $6*10^{-3}N$$\\$ Charge on the first sphere,$q_1=2*10^{-7}C$$\\$ Charge on the second sphere, $q_2 = 3 * 10^{-7} C$$\\$ Distance between the spheres, $r=30 cm =0.3 m$$\\$ Electrostatic force between the spheres is given by the relation,$\\$ $F=\dfrac{q_1q_2}{4 \pi \in_0r^2}$$\\$ Where, $\in _0 =$ Permittivity of free space$\\$ $\dfrac{1}{4 \pi \in_0}=9*10^9Nm^2C^{-2}\\ F=\dfrac{9*10^9*2*10^{-7}*3*10^{-7}}{(0.3)^2}=6*10^{-3 } N$$\\$ Hence, force between the two small charged spheres is $6*10^{-3}N$. The charges are of same nature. Hence, force between them will be repulsive.

2   The electrostatic force on a small sphere of charge $0.4 \mu C$ ue to another small sphere of charge $- 0.8\mu C$ in air is $0.2 N.$$\\$ (a) What is the distance between the two spheres?$\\$ (b) What is the force on the second sphere due to the first?

Solution :

3   The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is $0.2 N.$$\\$ (a) What is the distance between the two spheres?$\\$ (b) What is the force on the second sphere due to the first?

Solution :

(a) Electrostatic force on the first sphere, $F = 0.2 N$$\\$ Charge on this sphare,$q_1=0.4 \mu C=0.4 * 10 ^{-6}C$$\\$ Charge on the second sphare,$q_2=-0.8 \mu C=-0.8 \mu C=-0.8*10^{-6}C$$\\$ Electrostatic force between the spheres is given by the relation,$\\$ $F=\dfrac{q_1q_2}{4 \pi \in_0r^2}$$\\$ Where,$\in_0=$ Permittivity of free space$\\$ And,$\dfrac{1}{4 \pi \in_0}=9*10^9 n m^2 C^{-2}$$\\$ $r^2=\dfrac{q_1q_2}{4 \pi \in_0F}\\ =\dfrac{0.4*10^{-6}*8*10^{-6}*9*10^9}{0.2}\\ =144*10^{-4}\\ r=\sqrt{144*10^{-4}}=0.12m $$\\$ The distance between the two spheres is $0.12 m.$$\\$ (b) Both the spheres attract each other with the same force (according to Newton’s $3^{ rd}$ Law). Therefore, the force on the second sphere due to the first is $0.2 N.$

4   Check that the ratio $ke^2 / G m_e m_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution :

The given ratio is $\dfrac{ke^2}{Gm_em_p}$$\\$ Where,$\\$ $G=$Gravitational constant$\\$ Its unit is $Nm^2 kg^{-2}.$$\\$ $m_e $ and $m_p$=Masses of electron and proton.$\\$ Their unit is kg.$\\$ $e$ = Electric charge$\\$ Its units is C.$\\$ $k=A$ constant$\\$ $=\dfrac{1}{4 \pi \in_0}$$\\$ $\in_0=$ Permittivity of free space$\\$ Its unit is $nm^2C^{-2}$$\\$ Therefore, unit of the given ratio $\dfrac{ke^2}{Gm_em_p}$$\\$ $=\dfrac{[N m^2 C^{-2}][C^2]}{[Nm^2 kg^{-2}][kg][kg]}$$\\$ Hence, the given ratio is dimensionless.$\\$ $e=1.6*10^{-19}C\\ G=6.67*10^{-11} Nm^2 kg^{-2}\\ m_e=9.1*10^{-31}kg\\ m_p=1.66*10^{-27}kg$$\\$ Hence, the numerical value of the given ratio is$\\$ $\dfrac{ke^2}{Gm_em_p}=\dfrac{9*10^9*(1.6*10^{-19})^2}{6.67*10^{-11}*9.1*10^{-3}*1.67*10^{-22}} \approx 2.3*10^{39}$$\\$ This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

5   (a) Explain the meaning of the statements ‘electric charge of a body is quantized’.$\\$ (b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?

Solution :

(a) Electric charge of a body is quantized. This means that only integral $(1, 2,....., n )$ number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.$\\$ (b) In macroscopic or large scale charges, the number of charges are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

6   When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution :

Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

7   Four point charges $q_A=2\mu C, q_B=-5\mu C, q_c = 2\mu C ,$ and $q _D = 5\mu C$ are located at the corners of a square $ABCD $ of side $10 cm$. What is the force on a charge of $1\mu C$ placed at the centre of the square?

Solution :

The given figure shows a square of side $10 cm$ with four charges placed at its corners. $O$ is the centre of the square.$\\$ Where,$\\$ (Sides)$AB=BC=CD=AD=10 cm$$\\$ (Diagonals)$AC=BD=10\sqrt{2}cm$$\\$ $AO=OC=DO=OB=5\sqrt{2}cm$$\\$ A charge of amount $1\mu C$ is placed at point $O$. Force of repulsion between charges placed at corner $A$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner $C$ and centre $O$. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner $B$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner $D$ and centre $O$. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corners of the square, on $1\mu C$ charge at centre $O$ is zero.

8   (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?$\\$ (b) Explain why two field lines never cross each other at any point?

Solution :

(a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.$\\$ (b) If two field lines cross each other at a point, then electric field intensity will show two different directions at that point, as two different tangents can be drawn at the point of intersection and they represent the direction of electric field intensity at that point. This is not possible. Hence, two field lines never cross each other.

9   Two point charges $q_A = 3\mu C$ and $q_B =-3\mu C$ are located $20 cm$ apart in vacuum. (a) What is the electric field at the midpoint $O$ of the line $AB $ joining the two charges? $\\$ (b) If a negative test charge of magnitude $1.5*10^{-9}C $ is placed at this point, what is the force experienced by the test charge?

Solution :

[Since the values of $E_1$ and $E_2$ are same, the value is multiplied with $2$]$\\$ $=5.4*10^6 N/C \text{along} OB$$\\$ Therefore, the electric field at mid-point $O$ is $=5.4*10^6 NC^{-1} \text{along } OB$$\\$ (b)A test charge of amount $1.5*10^{-9}C$ is placed at mid-point $O$. $q=1.5*10^9 C$$\\$ Force experienced by the test charge = F$\\$ $\therefore F=qE\\ 1.5*10^{-9}*5.4*10^6\\ 8.1*10^{-3}N$$\\$ The force is directed along line $OA$. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Therefore, the force experienced by the test charge is $= 8.1 * 10 ^{- 3} N \text{along} OA.$

(a) The situation is represented in the given figure. $O$ is the mid-point of line $AB$.$\\$ Distance between the two charges, $AB = 20 cm$$\\$ $\therefore AO = OB = 10 cm$$\\$ Net electric field at point $O = E$$\\$ Electric field at point $O$ caused by $+3\mu C$ charge,$\\$ $E_1=\dfrac{3*10^{-6}}{4 \pi \in_0(AO)^2}\\ \dfrac{3*10^{-6}}{4 \pi \in_0(10*10^{-2})} N/C \text{alomg}OB$$\\$ Where,$\\$ $\in_0=$ Permittivity of free space$\\$ $\dfrac{1}{4\pi \in_0}=9*10^9 Nm^2 C^{-2}$$\\$ Magnitude of electric field at point $O$ caused by $-3\mu C$ charge,$\\$ $ E_2=|\dfrac{-3*10^{-6}}{4 \pi \in_0(OB)^2}| \\ =\dfrac{3*10^{-6}}{4 \pi \in_0(10 * 10^{-2})^2} N/c \text{along} OB\\ \therefore E=E_1+E_2\\ =2*[(9*10^9)*\dfrac{3*10^{-6}}{(10*10^{-2})^2}]$$\\$

10   A system has two charges $q_A = 2.5*10^{-7} C$ and $q_B=- 2.5*10^{-7} C$ located at points $A :(0,0, - 15cm )$ and $B : ( 0,0, +15cm )$ , respectively. What are the total charge and electric dipole moment of the system?

Solution :

Both the charges can be located in a coordinate frame of reference as shown in the given figure.$\\$ At A, amount of charge, $ q_A=2.5*10 ^{-7}C$$\\$ At B, amount of charge , $ q_B=-2.5*10^{-7}C$$\\$ Total charge of the system,$\\$ $q=q_A +q_B\\ =2.5*10^7C-2.5*10^{-7}C\\ =0$$\\$ Distance between two charges at points A and B,$\\$ $d=15+15=30 cm =0.3 m$$\\$ Electric dipole moment of the system is given by,$\\$ $p=q_A*d=q_B *d\\ =2.5*10^{-7}*0.3\\ =7.5*10^{-8 }C m$ along negative z-axis$\\$ Therefore, the electric dipole moment of the system is $7.5*10^{-8}C m$ along negative Z- axis.