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Question

A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is ρ. An electron (charge e , mass m) is released inside the cavity from point P as shown in figure. The center of sphere and center of cavity are seperated by a distance a. The time after which the electron again touches the sphere is


A
62 ε0meρ
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B
2 ε0meρ
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C
6 ε0meρ
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D
ε0meρ
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Solution

The correct option is A 62 ε0meρ
Given,

Radius of cavity =r



Electric field inside the cavity with respect to center of sphere , E=ρa3ε0 ....(1)
Since, acceleration of an electron A=eEm ....(2)
Using (1) in (2) we get,'

A=eρa3ε0m ( in backward direction )

Distance travelled by electron where it hits the wall of cavity is
S=2rcosθ
Substituting the value of (θ) we get,

S=2rcos45=r2

Since, the acceleration is constant, we can use the kinematic relation
S=ut+12At2
From the data given in the question,

S=12At2

t=2SA

Substituting the data obtained in the above equation we get,

t=62 rε0mepa

Hence, option (a) is the correct answer.


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