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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 √6√2 ε0meρGiven, Radius of cavity =r Electric field inside the cavity with respect to center of sphere , →E=ρ→a3ε0 ....(1) Since, acceleration of an electron →A=e→Em ....(2) Using (1) in (2) we get,' →A=eρ→a3ε0m ( 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∘=r√2 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=√6√2 rε0mepa Hence, option (a) is the correct answer.

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