Question

A thin spherical insulating shell of radius $ R $ carries a uniformly distributed charge such that the potential at its surface is $ {{V}_{0}} $ ​. A hole with a small area $ \alpha 4\pi {{R}^{2}}\left( \alpha \langle \langle 1 \right) $ is made on the shell without affecting the rest of the shell. Which one of the following statements is found to be correct?
A. the ratio of the potential at the centre of the shell to that of the point at $ \dfrac{1}{2}R $ from centre towards the hole will be $ \dfrac{1-\alpha }{1-2\alpha } $
B. the potential at the centre of shell is reduced by \[2\alpha {{V}_{0}}\]
C. the magnitude of electric field at the centre of the shell is reduced by \[\dfrac{\alpha {{V}_{0}}}{2R}\]
D. the magnitude of electric field at a point, located on a line passing through a hole and shell’s centre, on a distance \[2R\] from the centre of the spherical shell will be reduced by \[\dfrac{\alpha {{V}_{0}}}{2R}\]

Answer 1

Hint: Gauss’s law can be helpful in solving this question. Gauss's law is also called Gauss's flux theorem. This law is related to distribution of electric charge to the net electric field. The surface under consideration should be a closed one which is having a volume like a spherical surface.

Complete step-by-step answer:

We know that the potential at a surface is given as,
 $ {{V}_{0}}=\dfrac{KQ}{R} $ \[{{E}_{C}}=\dfrac{K\alpha Q}{{{R}^{2}}}=\dfrac{\alpha {{V}_{0}}}{R}\]
And also the potential at the centre C will be,
 $ {{V}_{C}}=\dfrac{KQ}{R}-\dfrac{K\alpha Q}{R}={{V}_{0}}\left( 1-\alpha \right) $
Potential at the point B will be,
 $ {{V}_{B}}=\dfrac{KQ}{R}-\dfrac{K\alpha Q}{\dfrac{R}{2}}={{V}_{0}}\left( 1-2\alpha \right) $
Taking the ratio between these two can be written as,
 $ \dfrac{{{V}_{c}}}{{{V}_{B}}}=\dfrac{\left( 1-\alpha \right)}{\left( 1-2\alpha \right)} $
The electric field acting at the point A is,
\[{{E}_{A}}=\dfrac{KQ}{{{\left( 2R \right)}^{2}}}-\dfrac{K\alpha Q}{{{R}^{2}}}=\dfrac{KQ}{4{{R}^{2}}}-\dfrac{\alpha {{V}_{0}}}{R}\]
That is the electric field will be reduced by\[\dfrac{\alpha {{V}_{0}}}{R}\],
The electric field experienced at C is given as,
\[{{E}_{C}}=\dfrac{K\alpha Q}{{{R}^{2}}}=\dfrac{\alpha {{V}_{0}}}{R}\]
That is, the electric field at C is increasing by\[\dfrac{\alpha {{V}_{0}}}{R}\].

So, the correct answer is “Option A”.

Note: Electric field is described as the electric force experiencing per unit charge. The direction of the field is the same as the direction of the force it would exert on a positive test charge. The electric field is radially outward in the case of a positive charge and radially inwards for a negative point charge.