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 $$\alpha 4\pi {R^2}\left( {a < < 1} \right)$$ is made on the shell without affecting the rest of the shell. Which one of the following statements is correct ?
A.
The magnituide of electric field at a point, located on a line passing through the hole and shell’s center, on a distance $$2R$$ from the centre of the spherical shell will be reduced by $$\frac{{\alpha {V_0}}}{{2R}}$$
B.
The ratio of the potential at the centre of the shell to that of the point at $$\frac{1}{2}R$$ from the centre towards the hole will be $$\frac{{1 - \alpha }}{{1 - 2\alpha }}$$
C.
The magnitude of electric field at the centre of the shell is reduced by $$\frac{{\alpha {V_0}}}{{2R}}$$
D.
The potential at the centre of the shell is reduced by $${2\alpha {V_0}}$$
Answer :
The ratio of the potential at the centre of the shell to that of the point at $$\frac{1}{2}R$$ from the centre towards the hole will be $$\frac{{1 - \alpha }}{{1 - 2\alpha }}$$
Solution :
Let $$Q$$ be the total charge on the sphere. Then surface charge density is $$\frac{Q}{{4\pi {R^2}}}.$$ A hole is now cut of area $$\alpha \left( {4\pi {R^2}} \right).$$ The charge on this hole is
$$\eqalign{ & \frac{q}{{\alpha \left( {4\pi {R^2}} \right)}} = \frac{Q}{{4\pi {R^2}}} \cr & \therefore q = \alpha Q\,\,\,\,\,\,{\text{Also}}\,{V_0} = \frac{{KQ}}{R}. \cr} $$
Now we visualise this situation as a complete spherical distribution of positive charge on the surface with a negative charge of the same surface charge density on the hole. This negative charge can be treated as a point charge.
Potential
$$\eqalign{ & {V_P} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{R} = \frac{{KQ}}{R}\left( {1 - \alpha } \right) \cr & = {V_0}\left( {1 - \alpha } \right) = {V_0} - \alpha {V_0} \cr} $$
Therefore option (D) is incorrect.
$$\eqalign{ & {V_A} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{{\frac{R}{2}}} = \frac{{KQ}}{R}\left( {1 - 2\alpha } \right) \cr & \therefore \frac{{{V_P}}}{{{V_A}}} = \frac{{1 - \alpha }}{{1 - 2\alpha }} \cr} $$
Therefore option (B) is correct.
Electric field
$$\eqalign{ & {\left( {{E_B}} \right)_{initial{\text{ }}}} = \frac{{K{Q_0}}}{{{{\left( {2R} \right)}^2}}} \cr & {\left( {{E_B}} \right)_{final{\text{ }}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \frac{{K(\alpha Q)}}{{{R^2}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \alpha {V_0} \cr} $$
Option (A) is incorrect.
$$\eqalign{ & {\left( {{E_P}} \right)_{initial{\text{ }}}} = 0 \cr & {\left( {{E_P}} \right)_{final{\text{ }}}} = \frac{{K\left( {\alpha Q} \right)}}{{{R^2}}} = \frac{{\alpha {V_0}}}{R} \cr} $$
Option (C) is incorrect.
Let $$Q$$ be the total charge on the sphere. Then surface charge density is $$\frac{Q}{{4\pi {R^2}}}.$$ A hole is now cut of area $$\alpha \left( {4\pi {R^2}} \right).$$ The charge on this hole is
$$\eqalign{ & \frac{q}{{\alpha \left( {4\pi {R^2}} \right)}} = \frac{Q}{{4\pi {R^2}}} \cr & \therefore q = \alpha Q\,\,\,\,\,\,{\text{Also}}\,{V_0} = \frac{{KQ}}{R}. \cr} $$
Now we visualise this situation as a complete spherical distribution of positive charge on the surface with a negative charge of the same surface charge density on the hole. This negative charge can be treated as a point charge.
Potential
$$\eqalign{ & {V_P} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{R} = \frac{{KQ}}{R}\left( {1 - \alpha } \right) \cr & = {V_0}\left( {1 - \alpha } \right) = {V_0} - \alpha {V_0} \cr} $$
Therefore option (D) is incorrect.
$$\eqalign{ & {V_A} = \frac{{KQ}}{R} - \frac{{K\alpha Q}}{{\frac{R}{2}}} = \frac{{KQ}}{R}\left( {1 - 2\alpha } \right) \cr & \therefore \frac{{{V_P}}}{{{V_A}}} = \frac{{1 - \alpha }}{{1 - 2\alpha }} \cr} $$
Therefore option (B) is correct.
Electric field
$$\eqalign{ & {\left( {{E_B}} \right)_{initial{\text{ }}}} = \frac{{K{Q_0}}}{{{{\left( {2R} \right)}^2}}} \cr & {\left( {{E_B}} \right)_{final{\text{ }}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \frac{{K(\alpha Q)}}{{{R^2}}} = \frac{{KQ}}{{{{\left( {2R} \right)}^2}}} - \alpha {V_0} \cr} $$
Option (A) is incorrect.
$$\eqalign{ & {\left( {{E_P}} \right)_{initial{\text{ }}}} = 0 \cr & {\left( {{E_P}} \right)_{final{\text{ }}}} = \frac{{K\left( {\alpha Q} \right)}}{{{R^2}}} = \frac{{\alpha {V_0}}}{R} \cr} $$
Option (C) is incorrect.
