NCERT Solution : Electric Charges & Fields

Q1. What is the force between two small charged spheres having charges 2 × 10-7 C  and 3 × 10-7 C placed 30cm apart in air ?

Sol. Given q1 = 2 × 10-7 C , q2 = 3 × 10-7 C ,

r = 30 cm = 0.3 m , F = ?

By formula ,$ \displaystyle F = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{r^2} $

$ \displaystyle F = (9\times 10^9) \frac{2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^2} $

= 6 × 10-3 N (repulsive)

Q2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC 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 ?

Sol. Given , q1 = 0.4 μC  , q2 = −0.8 μC

F12 = 0.2 N

(a) We know that

$ \displaystyle F = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{r^2} $

$ \displaystyle r^2 = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{F} $

$ \displaystyle r^2 = (9\times 10^9) \frac{0.4\times 10^{-6}\times 0.8\times 10^{-6}}{0.2} $

r2 = 144 × 10-4

and r = 12 × 10-2 m

r = 12 cm

(b) Force on second sphere due to first is same i.e. 0.2 N

Q3. Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify ?

Sol. An electron and proton have a charge of 1.6 × 10-19 C each and their masses are 9.1×10-31 kg and 1.67 ×10-27 kg .

Electrostatic force b/w electron and proton ,

$\displaystyle F_e = \frac{1}{4\pi \epsilon_0 }\frac{e .e}{r^2} $

Gravitational force b/w electron and proton ,

$ \displaystyle F_g = G\frac{m_e . m_p}{r^2} $

On dividing ,

$ \displaystyle \frac{F_e}{F_g} = \frac{e^2}{4\pi \epsilon_0 G m_e m_p} $

On putting the values ,

$ \displaystyle \frac{F_e}{F_g} \approx 2.4 \times 10^{39} $

The ratio is quite large . This shows that electrostatic forces are much stronger than gravitational forces .

A common example is the lifting of a paper by charged comb against the force of entire Earth on that paper .

Q:4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised ’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges ?

Sol. (a) Quantisation of electric charge : It is now well known fact that all charges occurring in nature are positive or negative integral multiple of a basic unit of electric charge which we take as the magnitude of the charge on an electron . Hence charge on an electron is -e and that on a proton happens to be +e , while charge on a neutron is zero . Any charged body will have ±ne  charge , where n is an integer . This fact is called the quantisation of electric charge .

(b) At the macroscopic level one deals with charges that are enormous compared to the magnitude of charge e . Since e = 1.6 × 10-19 C , a charge of magnitude , say 1μC contains something like 1013 times the electronic charge . At this scale , the fact that the charge can increase or decrease only in units of e is not very different from saying that charge can take continuous value . Thus at the macroscopic level , the  quantisation has no practical consequence and can be ignored .

Q: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.

Sol. Charged is neither created nor destroyed . It is merely transferred from one body to another . Electrons are transferred from glass to silk , so glass has positive charge and silk has negative charge .

Q:6. Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC 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 ?

Sol. The center O of square is at equal distance of 10/√2 cm from each corners . Since opposite corners have equal charges , forces along both diagonals will be balanced . Resultant force on 1 μC charge at O will be zero .

Q:7. (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 ?

Sol. (a) They start from a positive charge and end at a negative charge . They are continuous , because force is continuous . They do not have sudden breaks , otherwise a moving test charge will have  to take jumps .

(b) Two lines of force do not intersect each other . If they intersect at a point , there will be two directions of field at that point . Since it is impossible , hence they don’t intersect .

Q: 8. Two point charges qA = 3 μC and qB = -3 μC 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 ?

Sol: qA = 3 μC = 3 × 10-6 C and qB = -3 μC = -3 × 10-6 C

(a) E0 = ?,

q0 = 1.5 × 10-9 C ,

Electrostatics

E at O due to charge at A,

$ \displaystyle E_A = \frac{1}{4\pi \epsilon_0}\frac{q_A}{r_A^2} $

$ \displaystyle E_A = (9\times 10^9) \frac{3\times 10^{-6}}{(0.1)^2} $

= 27 × 105 N/C along OB

Also, E at O due to charge at B,

EB = 27 × 105 N/C along OB

Resultant E at O ,

EO = EA + EB

= 27 × 105 + 27 × 105

= 54 × 105 N/C along OB

(b) For point O, E = 54 × 105 N/C , q0 = 1.5 × 10-9 C ,

By relation, F = q0 E

F = 1.5 × 10-9 × 54 × 105

= 81 × 10-4

= 8.0 × 10-3 N along OA.

Q9. A system has two charges qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system ?

Sol. The system is an electric dipole of charge strength q = (2.5 × 10–7 −2.5 × 10–7)C = Zero

Electric dipole moment ,

p = q × 2a = 2.5 × 10–7 × 0.30

= 7.5 × 10–8 C-m along Z-axis

Q10. An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1. Calculate the magnitude of the torque acting on the dipole.

Sol . Given p = 4 × 10–9 C- m

E = 5 × 104 NC–1 and θ = 30°

By Formula , τ = p E Sinθ

= 4 × 10–9 × 5 × 104  × 0.5 = 10–4 J

Q.11. A polythene piece rubbed with wool is found to have a negative charge of 3.2 × 10–7 C.
(a) Estimate the number of electrons transferred (from which to which ?)
(b) Is there a transfer of mass from wool to polythene ?

Sol. (a) Given , q = −3.2 × 10–7 C , e = −1.6 x 10-19 C

By relation ,  q = n e

$ \displaystyle n = \frac{q}{e} $

$ \displaystyle n = \frac{-3.2\times 10^{-7}}{-1.6\times 10^{-19}} $

= 2 × 1012

These electrons are transferred from wool to polythene .

(b) Since electrons have a definite mass (9.1 × 10−31 kg ) , the transfer of electrons may result in mass transfer by amount = 2 × 1012 × 9.1 × 10−31 kg = 18.2 × 10−19 kg which is negligible

Q12. (a) 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 × 10–7 C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved ?

Sol. (a)q1 = 6.5 × 10–7 C , q2 = 6.5 × 10–7 C

r = 50 cm = 0.5m , F = ?

By formula ,

$ \displaystyle F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2} $

$ \displaystyle F = ( 9\times 10^9 ) \frac{(6.5\times 10^{-7})^2}{(0.5)^2} $

F = 1.521 × 10−2 N

(b) Doubling charge on each sphere increases the force four times . Making the distance half , further increases the force four times .

So new force becomes 16 times .

Now F = 1.521 × 10-2 × 16 = 24.34 × 10-2 N

Q13. Suppose the spheres A and B in Exercise 1.12 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 ?

Sol. Before contact , sphere A has charge 6.5 × 10–7 C , while C has no charge . On contact , A and C both have equal charge , i.e. 3.25 × 10–7 C each .

Then sphere C (with charge 3.25 × 10–7 C ) comes in contact with sphere B has charge

$ \displaystyle = \frac{(3.25+6.5)\times 10^{-7}}{2} $

= 4.875 × 10–7 C

Now , q1 = 3.25 × 10–7 C

q2 = 4.875 × 10–7 C

$ \displaystyle F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2} $

$ \displaystyle F = (9\times 10^9) \frac{3.25\times 10^{-7} (4.875 \times 10^{-7})}{(0.5)^2} $

F = 0.5625 × 10-2 N

Q14. Figure  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 ?

Electrostatics

Sol.
Particle 1 and 2  have negative charge .

Particle 3 has positive charge .

Displacement $ \displaystyle y \propto \frac{q}{m}$

Particle 3 has highest charge to mass ratio , because its path is more curved .

Q15. Consider a uniform electric field E = 3 × 103 î N/C.

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?

(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis ?

Sol. Given , E = 3 × 103 î N/C

A = 10 × 10 = 100 cm2 = 10-2 m2

Since surface lies in Y-Z plane , normal to the surface is along X-axis , A = 10−2 î m2

(a) θ = 0° , φ = ?

(b) θ = 60° , φ = ?

By relation , φ = E A cos θ

Substituting  the values , we get

(a) φ = 3 × 103 × 10-2 × 1 = 30 Nm2/C

(b) φ = 3 × 103 × 10-2 × 0.5 = 15 Nm2/C

Next Page →