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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.