Electric charges and fields NCERT numerical

 

Class 12 Physics NCERT Solutions Electric Charges and Fields 

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Q 1.1) What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?

Soln: Given,

The Charge on the 1st sphere and 2nd sphere is q1 = 2 x 10-7 C and q2 = 3 x 10-7 C

The distance between two charges is given by r = 30cm = 0.3m

The electrostatic force between the spheres is given by the relation :

F = $\frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}$

Here,

$\epsilon _{o}$ = permittivity of free space and,

$\frac{1}{4\pi \epsilon _{o}}$ = 9 x 109 Nm2C-2

Force, F = $\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}$ = 6 x 10‑3 N.

The force between the charges will be repulsive as they have the same nature.

 

Q 1.2) The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the 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?

Soln:

(a) Given,

The charge on 1st sphere (q1 ) and 2nd sphere (q2) is 0.4 µC or 0.4 × 10-6 C and -0.8 × 10-6C respectively.

The electrostatic force on the 1st sphere is given by F = 0.2N.

Electrostatic force between the spheres is given by the relation :

F = $\frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}$

Here,

$\epsilon _{o}$ = permittivity of free space and,

$\frac{1}{4\pi \epsilon _{o}}$ = 9 × 109 Nm2C-2

r2 = $\frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{F}$

= $\frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^{9}}{0.2}$ = 144 x 10-4

r = $\sqrt{144 \times 10^{-4}}$ = 12 x 10-2 = 0.12m

Therefore, the distance between the two spheres = 0.12 m

 

(b) Since the spheres have opposite charges, the force on the second sphere due to the first sphere will also be equal to 0.2N.

 

Q 1.3) Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Soln.:

The ratio to be determined is given as follows :

$\frac{ke^{2}}{Gm_{e}m_{p}}$

where G is the gravitational constant in N m2 kg-2

me and mp is the masses of electron and proton in kg.

e is the electric charge (unit – C)

k = $\frac{1}{4\pi \epsilon _{o}}$  (unit – Nm2C-2)

Therefore, the unit of given ratio,

$\frac{ke^{2}}{Gm_{e}m_{p}}$ = $\frac{[Nm^{2}C^{-2}][C^{-2}]}{[Nm^{2}kg^{-2}][kg][kg]}$ = M0L0T0

So, the given ratio is dimensionless.

Given,

e = 1.6 x 10-19 C

G = 6.67 x 10-11 N m2 kg-2

me = 9.1 x 10-31 kg

mp = 1.66 x 10-27 kg

Putting the above values in the given ratio, we get

$\frac{ke^{2}}{Gm_{e}m_{p}}$ = $\frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}$ = 2.3 x 1039

So, the above ratio is the ratio of the electric force to the gravitational force between a proton and an electron when the distance between them is constant.

 

Q 1.4) (i) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(ii) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Soln.:

(i) The ‘electric charge of a body is quantized’ means that only integral (1, 2, …n)  numbers of electrons can be transferred from a body to another.

Charges cannot get transferred in fractions. Hence, the total charge possessed by a body is only in integral multiples of electric charge.

(ii) In the case of large scale or macroscopic charges, the charge which is used over there is comparatively too huge to the magnitude of the electric charge. Hence, on a macroscopic level, the quantization of charge is of no use Therefore, it is ignored and the electric charge is considered to be continuous.

Q 1.5) 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.

Soln.:

When two bodies are rubbed against each other,  a charge is developed on both bodies. These charges are equal but opposite in nature. And this phenomenon of inducing a charge is known as charging by friction. The net charge on both of the bodies is 0 and the reason behind it is that an equal amount of charge repels it. When we rub a glass rod with a silk cloth, charge with opposite magnitude is generated over there. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

 

Q 1.6) 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 µC placed at the centre of the square?

Soln.:

In the above picture, we have shown the square mentioned in the question. Whose side is 10 cm and four charges are placed at the corners of the squares And O is the centre of the square.

Where,

(Sides) AB = BC = CD = DA = 10 cm

(Diagonals) AC = BD = $10 \sqrt{2} \; cm$

AO = OC = DO = OB = $5 \sqrt{2} \; cm$

At the centre point ‘O’, we have placed a charge of $1 \; \mu C$

In the above case, the repulsive force between the corner A and the centre O is same in magnitude with the repulsive force by the corner C to the centre O, but these forces are opposite in direction. Hence, these forces will cancel each other and from A and C no forces are applied on the centre O. Similarly, from the corner C the attractive force is applying on to the centre O and another force with the same magnitude is applying on the centre O, also these two forces are opposite in direction hence they are also opposing each other.

Therefore, the net force applying to the centre is zero. Because all the forces here are being cancelled by each other.

 

Q 1.7)  (i) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(ii) Explain why two field lines never cross each other at any point.

 

Soln.:

(i) When a charge is placed in an electrostatic field then it experiences a continuous force. Therefore, an electrostatic field line is a continuous curve. And a charge moves continuously and does not jump from on point to the other. So, the field line cannot have a sudden break.

(ii) if two field lines will cross each other at any point then at that point the field intensity will start shooing two directions at the same point which is impossible. Therefore, two field lines can never cross each other.

Q 1.8) Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.

(i) What is the electric field at the midpoint O of the line AB joining the two charges?

(ii) 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?

Soln.:

(i) The figure given below shows the situation given to us, in which AB is a line and O is the midpoint.

Distance between two charges, AB = 20 cm

Therefore, AO = OB = 10 cm

Total electric field at the centre is (Point O) = E

Electric field at point O caused by $+ 3 \; \mu C$ charge,

$E _{1} = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( OA \right )^{2}} = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} NC ^{-1}$         along OB

Where $\epsilon _{0}$ = Permittivity of free space and $\frac{1}{ 4 \pi \epsilon _{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$

Therefore,

Electric field at point O caused by $- 3 \; \mu C$ charge,

$E _{2} = \left | \frac {1}{4 \pi \epsilon _{0}}. \frac{- 3 \times 10 ^{-6}}{\left ( OB \right )^{2}} \right | = \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} NC ^{-1}$                 along OB

$∴ E _{1} + E _{2} = 2 \times \frac {1}{4 \pi \epsilon _{0}}. \frac{3 \times 10 ^{-6}}{\left ( 10 \times 10 ^{-2} \right )^{2}} \; NC ^{-1}$              along OB

[Since the magnitudes of $E _{1} \; and \; E _{2}$are equal and in the same direction]

 

$∴ E = 2 \times 9 \times 10 ^{9} \times \frac{3 \times 10^{-6}}{\left (10 \times 10^{-2} \right ) ^{2}} \; NC ^{-1} \\ \\ = 5.4 \times 10^{6} NC ^{-1}$ along OB

Therefore, the electric field at mid – point O is  $5.4 \times 10^{6} NC ^{-1}$ along OB.

 

(ii) A test charge with a charge potential of $1.5 \times 10 ^{-9} \; C$ is placed at mid – point O.

$q = 1.5 \times 10 ^{-9} \; C$

Let the force experienced by the test charge be  F

Therefore, F = qE

= $1.5 \times 10 ^{-9} \times 5.4 \times 10 ^{6} \\ \\ = 8.1 \times 10 ^{-3} \; N$

The force is directed along line OA because the negative test charge is attracted towards point A and is repelled by the charge placed at point B. As a result, the force experienced by the test charge is $q = 8.1 \times 10 ^{-3} \; N$ along OA.

 

Q 1.9) A system has two charges $q _{A} = 2.5 \times 10 ^{-7} \;C \; and \; q _{B} = -2.5 \times 10 ^{-7}\;C$ located at points A :(0, 0, -15 cm) and B (0, 0, + 15 cm), respectively. What is the total charge and electric dipole moment of the system?

Soln.:

The charges which are located at the given points are shown in the co-ordinate system as:

At point A, total charge amount, $q _{A} = 2.5 \times 10 ^{-7} \;C$

At point B, total charge amount, $q _{B} =  – 2.5 \times 10 ^{-7} \;C$

Total charge of the system is, $q _{A} + q_{B}= 2.5 \times 10 ^{-7} \;C – 2.5 \times 10 ^{-7} \;C = 0$

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,

$p = q _{A} \times d = q_{B} \times d = 2.5 \times 10 ^{-7} \times 0.3\\ \\ = 7.5 \times 10 ^{-8} \; C$ m along positive z – axis

Therefore, the electric dipole moment of the system is  $7.5 \times 10 ^{-8} \; C$ m along positive z−axis.

 

Q 1.10) An electric dipole with dipole moment $4 \times 10^{-9}\;  C\, m$ is aligned at 30° with the direction of a uniform electric field of magnitude $5 \times 10^{4}\; N C^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Soln.:

Electric dipole moment, $p = 4 \times 10^{-9}\;C\,m$

Angle made by p with a uniform electric field, $\theta = 30^{\circ}$

Electric field, $E = 5 \times 10^{4}\; NC^{-1}$

Torque acting on the dipole is given by the relation,

$\tau = pE \sin \theta\\ \\ = 4 \times 10^{-9} \times 5 \times 10^{4} \times \sin 30 = 20 \times 10 ^{-5} \times \frac {1}{2} = 10^{-4}\;Nm$

Therefore, the magnitude of the torque acting on the dipole is $10^{-4}\;Nm$

 

Q 1.11) A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}\; C$.

(i) Estimate the number of electrons transferred (from which to which?)

(ii) Is there a transfer of mass from wool to polythene?

 

Soln.:

(i) Since the wool is positively charged and the polythene is negatively charged, so we can say that few amounts of electrons are transferred from wool to polythene.

Charge on the polythene, q = $3 \times 10^{-7}\; C$.

Amount of charge on an electron, $e = -1.6 \times 10^{-19}\; C$

Let number of electrons transferred from wool to polythene be n

So, by using the given equation we can calculate the value of n,

q = ne

$\Rightarrow n = \frac{q}{e} = \frac{-3 \times 10 ^{-7}}{-1.6 \times 10 ^{-19}} = 1.87 \times 10 ^{12}$

 

Therefore, the number of electrons transferred from wool to polythene is $1.87 \times 10 ^{12}$

 

(ii) Yes,

Mass is also transferred as an electron is transferred from wool to polythene and an electron particle have some mass.

Mass of an electron, $m_{e} = 9.1 \times 10 ^{-31} kg$

Total mass transferred , m = $m_{e} \times n \\ \\ = 9.1 \times 10 ^{-31} \times 1.87 \times 10^{12} \\ \\ = 1.701 \times 10^{-18} \; kg$

Here, the mass transferred is too low that it can be neglected.

 

Q 1.12) (i) Two insulated charged copper spheres A and B have their centres
separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$ each? The radii of A and B are negligible compared to the distance of separation.

(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

 

Soln.:

(i) Charge on sphere A, $q_{A} = 6.5 \times 10^{-7} \; C$

Charge on sphere B, $q_{B} = 6.5 \times 10^{-7} \; C$

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres, $F = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}}$

Where $\epsilon _{0}$ = Permittivity of free space and $\frac{1}{ 4 \pi \epsilon _{0}} = 9 \times 10^{9} \; Nm^{2} C^{2}$

Therefore,

$F = \frac{9 \times 10 ^{9} \times \left ( 6.5 \times 10^{-7} \right )^{2}}{\left ( 0.5 \right )^{2}}\\ \\ = 1.52 \times 10^{-2}\; N$

Therefore, the force between the two spheres is $1.52 \times 10^{-2}\; N$

 

(ii) After doubling the charge,

Charge on sphere A, $q_{A} = 1.3 \times 10^{-6} \; C$

Charge on sphere B, $q_{B} = 1.3 \times 10^{-6} \; C$

The distance between the spheres is halved.

$∴ r = \frac {0.5}{2} = 0.25 \; m$

Force of repulsion between the two spheres,

$F = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}} = \frac{9 \times 10 ^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6} }{\left ( 0.25 \right )^{2}}\\ \\ = 16 \times 1.52 \times 10^{-2} \\ \\ = 0.243 \; N$

Therefore, the force between the two spheres is 0.243 N.

 

Q 1.13) Suppose the spheres A and B in Exercise 12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally
removed from both. What is the new force of repulsion between A and B?

 Soln.:

Distance between the spheres, A and B, r = 0.5 m

Initially, the charge on each of sphere $q = 6.5\times 10^{-7} \; C$

When the sphere A is touched with An uncharged sphere C, then half of the charge will be transferred to the sphere C. Hence the charge on both the spheres A and C will be q/2.

After that when sphere C with charge q/2 is brought in touch with sphere B with charge q, then charge on each of the sphere will be divided in two equal parts, is.

$\frac{1}{2}\left ( q + \frac{q}{2} \right ) = \frac{3q}{4}$

Hence, charge on each of the spheres, C and B , is $\frac{3q}{4}$

Force of repulsion between sphere A and B is:

$F = \frac{1}{4 \pi \epsilon _{0}}. \frac{q _{A} q_{B}}{r ^{2}} = \frac{1}{4 \pi \epsilon _{0}}. \frac{\frac{q}{2} \times \frac{3q}{4}}{r^{2}} \\ \\ = \frac{1}{4 \pi \epsilon _{0}}. \frac{3q^{2}}{8r^{2}} \\ \\ = \frac{9 \times 10^{9} \times 3 \times \left ( 6.5 \times 10^{-7} \right )^{2}}{8 \times \left ( 0.5 \right )^{2}} = 5.703 \times 10^{-3} \; N$

Therefore, the force of attraction between the two spheres is $5.703 \times 10^{-3} \; N$

 

Q 1.14) The figure below shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Soln.:

We can see here that the particles 1 and 2 are moving in the direction of the positive charge and we know that the opposite charge attracts each other and the same charge repels each other. So, here we can say that the charged particles 1 and 2 which are going towards the charged particle are negatively charged. And the particle 3 is being attracted towards the negative charge. So, the particle 3 will be the positively charged particle.

The EMF or charge to mass ratio is directly proportional to the amount of depletion or depletion at a given velocity. Here we can see that the particle 3 is depleting more as compared to the other two. Therefore, it will have a higher charge to mass ratio.