Question

The dielectric strength of air at N.T.P. is $ 3 \times {10^6}{V \mathord{\left/
 {\vphantom {V m}} \right.} m} $ . Then the maximum charge that can be given to a spherical conductor of radius 3m is:
(A) $ 3 \times {10^{ - 1}}C $
(B) $ 3 \times {10^2}C $
(C) $ 3 \times {10^{ - 3}}C $
(D) $ 3 \times {10^{ - 4}}C $

Answer 1

Hint: We need to calculate the capacitance of the spherical conductor and calculate the potential difference from the product of dielectric strength and radius of the conductor. Calculate the net charge from the capacitance and potential difference of the spherical conductor.

Formula Used: The formulae used in the solution are given here.
The capacitance $ C = 4\pi { \in _0}R $ where $ R $ is the radius of the spherical conductor.
Capacitance $ C = {{{Q_{net}}} \mathord{\left/
 {\vphantom {{{Q_{net}}} V}} \right.} V} $ where $ {Q_{net}} $ is the charge on the capacitor and $ V $ is the potential difference between the ends of a conductor.
Potential Difference across the conductor, $ V = E \cdot R $ where $ E $ is the dielectric strength of air.

Complete step by step answer:
It has been given that, the dielectric strength of air at N.T.P. is $ 3 \times {10^6}{V \mathord{\left/
 {\vphantom {V m}} \right.} m} $ . A spherical conductor has radius 3m.
Thus, we can write, radius $ R = 3m $ .
Capacitance of the conductor is the ratio of the amount of electric charge stored on a conductor to a difference in electric potential. Capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential. Capacitance also implies an associated storage of electrical energy. If electric charge is transferred between two initially uncharged conductors, both become equally charged, one positively, the other negatively, and a potential difference is established between them.
The capacitance $ C = 4\pi { \in _0}R $ .
Now we know that the value of $ 4\pi { \in _0} $ is equal to, $ \dfrac{1}{{9 \times {{10}^9}}} $ .
 $ C = \dfrac{1}{{9 \times {{10}^9}}} \times 3F $
 $ \Rightarrow C = \dfrac{1}{3} \times {10^{ - 9}}F $
Now, capacitance is also equal to, $ C = {{{Q_{net}}} \mathord{\left/
 {\vphantom {{{Q_{net}}} V}} \right.} V} $ where $ {Q_{net}} $ is the charge on the capacitor and $ V $ is the potential difference between the ends of a conductor.
Now, $ V = E \cdot R $ where $ E $ is the dielectric strength of air.
Substituting the values given in the question, $ E = 3 \times {10^6}{V \mathord{\left/
 {\vphantom {V m}} \right.} m} $ and $ R = 3m $ , we get,
 $ V = 3 \times {10^6} \times 3 $
 $ \Rightarrow V = 9 \times {10^6}V $
The net charge is given by, $ {Q_{net}} = C \times V $
 $ {Q_{net}} = \dfrac{1}{{3 \times {{10}^9}}} \times 9 \times {10^6} $
 $ \Rightarrow {Q_{net}} = 3 \times {10^{ - 3}}C $
Hence, the correct answer is Option C.

Note:
Dielectric strength is defined as the electrical strength of an insulating material. In a sufficiently strong electric field the insulating properties of an insulator breaks down allowing flow of charge. Dielectric strength is measured as the maximum voltage required to produce a dielectric breakdown through a material. It is expressed as Volts per unit thickness.