A spherical shell of radius R has a uniformly distributed charge ,then electric field varies as

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