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A uniformly charged solid shpere fo rad...

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

A

`R_1=0 and R_2 gt (R_4 - R_3)`

B

`R_(1) ne 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 gtR_4`

Text Solution

Verified by Experts

The correct Answer is:
C
Potential at the surface of the charged sphere
`V_0=(kQ)/R`
`implies V=(KQ)/r, r ge R`
`=(kQ)/(2R^3) (3R^2-r^2)` for `r le R`
`V_(centre) =V_c=(KQ)/(2R^3) times 3R^2=(3kQ)/(2R) =(3V_0)/2 implies R_1=0`
As potential decreases for outside points. Thus, according to the question, we can write
`V_R2=(5V_0)/4=(kQ)/(2R^3) (3R^2-R_2^2)`
`(5V_0)/R=V_0/(2R^2) (3R^2-R_2^2) or 5/2=3-(R_2/R)^2 or (R_2/R)^2=3-5/2=1/2 or R_2=R/sqrt2`
SImilarly
`V_(R_3) =(kQ)/R_3=(3V_0)/4 implies (KQ)/R_3=3/4 times (KQ)/R or R_3=4/3 R`
`V_(R_4)=(KQ)/(R_4)=V_0/4 implies (kQ)/R_4=1/4 times (KQ)/R or R_4=4R`
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