A uniformly charged solid shpere fo radius `R` has potential `V_(0)` (measured with respect to `oo`) on its surface. For this sphere the equipotentail surfaces with potentials `(3V_(0))/(2) , (5 V_(0))/(4), (3V_(0))/(4)` and `(V_(0))/(4)` have radius `R_(1), R_(2), R_(3)` and `R_(4)` respecatively. Then

Options

A. `R_(1) = 0` and `R_(2) gt (R_(4) - R_(3))`
B. `R_(1) != 0` and `R_(2) gt (R_(4) - R_(3))`
C. `R_(1) = 0` and `R_(2) lt (R_(4) - R_(3))`
D. `2R lt R_(4)`

Answer

C, D

Explanation

Here, `V_(0) = (kQ)/(R )` where `k = (1)/(4pi in_(0))`
When `r gt R, V_(r gt R) = (kQ)/(r )`
When `r lt R, V_(rltR) = (kQ)/(2 R^(3)) (3R^(2) - r^(2))`
At centre, `r = 0, V_("centre") = (3)/(2) (kQ)/(r ) = (3)/(2) V_(0)`
When `r = R_(2)`,
`V_(R_(2)) = (5 V_(0))/(4) = (5)/(4) (kQ)/(R ) = (kQ)/(2R^(3)) (3R^(2) - R_(2))` `V_(R_(2))` On solving, `R_(2) = (R )/(sqrt(2))`
When `r = R_(3)`,
`V_(R_(3)) = (3V_(0))/(4) = (3)/(4) (kQ)/(R ) = (kQ)/(R_(3))` or `R_(3) = (4)/(3) R`
When `r = R_(4), V_(R_(4)) = (V_(0))/(4) = (1)/(4) (kQ)/(R ) = (kQ)/(R_(4))`
or `R_(4) = 4R, i.e., R_(4) gt 2R`.