A Positive charge Q is placed on a conducting spherical shell with inner radius `R_(1)` and outer radius ` R_2` . A particle with charge q is placed at the center of the spherical cavity . The magnitude of the electric field at a point in the cavity , a distance r from center is

Options

A. zero
B. `(Q) /(4 pi epsi _(0)r^2)`
C. `(q) /( 4pi epsi _0 r^2)`
D. `((q+Q))/(4pi epsi _0 r^2)`

Answer

B

Explanation

total charge on shell = + Q
Here total charge on shell is + Q . So the ( -Q) is on the inner surface of shell
Hence , electric field inside the conductor = 0
Accoding the Gauss' s law
`oint Eds =(Q )/(epsi_0)`
` E oint ds = (Q)/( epsi _0)`
`E ( 4 pi r^2 )=(Q ) /( epsi_0)`
electric field `E= (Q) /( 4 pi r^2 epsi_0)`