A thin spherical indulating shell of radius R caries a uniformly distributed charge such that the potential act its surface is `V_(0)`. A hole with small area `alpha4piR^(2)(alpha lt lt 1)` is made in the shell without effecting the rest of the shell. Which one of the following is correct.

Options

A. The potential at the centre of the shells is reduced by `2 alpha V_(0)` .
B. The magnitude of electric field apoint located on a line passing through the hole and shell’s center, on a distance from the center of the spherical shell will be reduced by `(alphaV_(0))/(2R)`
C. The magnitude of electric field at the center of the shell is reduced by `(alpha V_(0))(2R)`
D. The ratio of the potential at the center of the cell to that of the point at `(1)/(2)R` from centertowards the hole will be `(1- alpha)/(1- 2alpha)`

Answer

D

Explanation

(A) The potential is reduced by `dv= k (dq)/(R)`
Also, `v_(0)=(kQ)/(R) therefore (dv)/(v_(0))=(dq)/(Q)=(sigma (alpha 4pi R^(2))/(sigma(4pi R^(2))=alpha rArr dv=alphav_(0)`
`therefore` A is wrong
(B) The field is reduced by `dE=(kdq)/(R^(2))=(dv)/(R=(alpha v_(0))/(R) therefore` B is wrong
(C) Initial field at centre is zero, so field increases by `dE=(kdq)/(R^(2))=(dv)/(R)=(alphav_(0))/(R) therefore` (C) is wrong
Potential at centre `=v_(0) -dv=(l-alpha)v_(0)`
Potential at `R/2=v_(0)-(kdq)/((R//2)) =v_(0) -2dv =(l-2alpha) v_(0) therefore " Ratio"=(1-alpha)/(1-2alpha)` (D) is correct