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A solid non-conducting sphere of radius ...

A solid non-conducting sphere of radius R is charged with a uniform volume charge density `rho`. Inside the sphere a cavity of radius r is made as shown in the figure. The distance between the centres of the sphere and the cavity is a. An electron of charge 'e' and mass 'm' is kept at point P inside the cavity at angle `theta = 45^(@)` as shown. If at t = 0 this electron is released from point P, calculate the time it will take to touch the sphere on inner wall of cavity again.

A

`sqrt((3Rm epsilon_(0)r)/(e rho a))`

B

`sqrt((6m epsilon_(0)r)/(e rho a))`

C

`sqrt((6sqrt(3) m epsilon_(0)R)/(e rho a))`

D

`sqrt((6sqrt(2)m epsilon_(0)r)/(e rho a))`

Text Solution

Verified by Experts

The correct Answer is:
D
The magnitude of electric field inside the cavity `E_("cavity")=(rhoa)/(3in_(0))`. It is directed along a in radially outward direction
For touching the sphere again, electron must move a distance `2rcos45^(@)` and time taken by electron for this is given as

`t=sqrt((2l)/(a))`
`impliessqrt((2(sqrt2r))/(eE//m))`
`impliest=sqrt((6(sqrt2mrin_(0)))/(erhoa))`
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