Electrostatics Question 258
A piece of cloud having area $ 25\times 10^{6}m^{2} $ and electric potential of $ 10^{5} $ volts. If the height of cloud is $ 750m $ , then energy of electric field between earth and cloud will be [RPET 1997]
Options:
V
2V
4V
D) ? 2V
Show Answer
Answer:
Correct Answer: A
Solution:
In case of a charged conducting sphere, the electric field inside the sphere is zero and the charge resides on the surface
$ {V _{\text{inside}}}={V _{\text{centre }}}={V _{\text{surface}}}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{q}{R} $
$ {V _{\text{outside}}}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{q}{r} $
If a and b are the radii of sphere and spherical shell respectively, then potential at their surface will be $ {V _{\text{sphere }}}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q}{a} $ and $ {V _{\text{shell}}}=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{Q}{b} $
$ \therefore $
$ V={V _{\text{sphere }}}-{V _{\text{shell}}}=\frac{1}{4\pi {\varepsilon _{0}}} \left[ \frac{Q}{a}-\frac{Q}{b} \right] $
Now when the shell is given charge (3Q), then the potential will be
$ V_{\text{sphere}}=\frac{1}{4\pi {\varepsilon _{0}}}\left[ \frac{Q}{a}+\frac{-3Q}{b} \right], $
$ V{’ _{\text{shell}}}=\frac{1}{4\pi {\varepsilon _{0}}}[ \frac{Q}{b}+\frac{(-2Q)}{b} ] $
$ \therefore $
$ V{’ _{\text{sphere }}}-V{’ _{\text{shell}}}=\frac{1}{4\pi {\varepsilon _{0}}}[ \frac{Q}{a}-\frac{Q}{b} ]=V $
BETA
