A uniformly charged solid sphere of radius R has potential `V_(0)` (measured with respect to `oo`) on its surface. For this sphere the equipotential surfaces with potentials `(3V_0)/(2),(5V_0)/(4), (3V_0)/(4) and (V_0)/(4)` have radius `R_(1), R_(2), R_(3) and R_(4)` respectively. Then

Options

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

Answer

C

Explanation

`V_(0) = 1/(4 pi epsilon_(0))q/R`[ q = charge of the sphere]
Now, `(3V_0)/4, (V_0)/4 lt V_(0) `,
So, these equipotential surfaces will be situated outside the sphere,
`R_(3) , R_(4) gt R`
In that case,
`(3V_0)/4 = 1/(4 pi epsilon_(0))q/(R_3) or, 3/4 1/(4 pi epsilon_(0)) q/R = 1/(4 pi epsilon_0) q/(R_3)`
i.e., `R_(3) = 4/3 R`
Again, `(V_0)/4 = 1/4 1/(4pi epsilon_0) q/R = 1/(4pi epsilon_0) q/(R_4) i.e., R_(4) = 4R`
`:. R_(4) - R_(3) = 8/3 R`
On the other hand, `(3V_0)/2, (5V_0)/4 gt V_(0)` , these equipotential surface are situated inside the sphere , `R_(1) , R_(2) lt R`.
Potential at any point inside the sphere at a distance r (r lt R) from the centre of the sphere,
`V = 1/(4pi epsilon_0) q/(2R^2) (3R^(2)-R_(2)^(2)) = (V_0)/(2R^2)(3R^(2)r^(2))`
`:.` In the first case,
`3/2 V_(0) = (V_0)/(2R^2)(3R^(2)-R_(1)^(2)) or, R_(1) = 0`,
In the second case,
`5/4 V_(0) = (V_0)/(2R^2)(3R^(2)-R_(2)^(2)) or, R_(2) = 1/(sqrt2)R` .