A solid sphere of radius `R` has total charge `2Q` and volume charge density `p=kr` where `r` is distance from centre. Now charges `Q` and `-Q` are placed diametrically opposite at distance `2a` where a is distance form centre of sphere such that net force on charge `Q` is zero then relation between a and `R` is

Options

A. `a=2^(-1//4)R`
B. `a=R/(sqrt(3))`
C. `a=8^(-1//4)R`
D. `a=(3R)/(2^(1//4)`

Answer

C

Explanation

Consider force on A due to B
`|F|=(Q^(2))/(4piepsilon_(0)(2a)^(2))`
`=(Q^(2))/(16 pi epsilon_(0)a^(2))`
Since A is in equilibrium, force on it due to changed sphere, `F_(s)=F` (magnitude) So, electric field at distance a from centre
`E=Q/(16b pi epsilon_(0)a^(2))`
If `r gtR`, result for electric field is `E=(2Q)/(1 pi epsilon_(0)r^(2))`
so assuming `agtR`
`E=Q/(16 pi epsilon_(0)a^(2))=(2Q)/(4 pi epsilon_(0)r^(2))`
`impliesr=2sqrt(2)aimpliesa=v/(2sqrt(v))`
But the result was only valid for `rgtR`
`implies` The point in inside the sphere and `a ltR`
Let electric field inside the sphere at distance r from centre b E, then by guass law
`E(4pir^(2))=1/(epsilon_(0))int_(0)^(r)kx(4pix^(2))dx`
`impliesE=(kr)^(2))/(4 epsilon_(0))`
Also `int_(0)^(R)kx(4pix^(2))dx=2Qimpliesk=(2Q)/(piR^(4))`
`impliesE=(Qr^(2))/(2pi epsilon_(0)R^(4))=(Qa^(2))/(2piepsilon_(0)R^(4))`
Comparing with `E=Q/(16 pi epsilona^(2))impliesa=8^(-1//4)R`