Consider the previous problem , let the outer shell have the charge ` - 4 q`. As in the above problem , the inner shell has the charge ` + 2q`. Calculate the electric field in terms of `q` and the distance `r` from the common center of the two shells for
`c lt r lt d`

Options

A. zero
B. `(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
C. `-(1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))`
D. `(1)/( 4 pi epsilon_(0)) ( 6 q)/( r^(2))`

Answer

A

Explanation

i. `r lt a. E = 0` , since charge enclosed is zero.
ii. `a lt r lt b. E = 0` , since charge enclosed is zero.
iii. `b lt r lt c. E = (1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))` , since charge enclosed is `+ 2q`.
iv. `c lt r lt d . E = 0` , since charge enclosed is zero ( Fig. S2.55).
v. `r gt d.E = - (1)/( 4 pi epsilon_(0)) ( 2q)/( r^(2))` , since charge enclosed is `- 2 q`.