1

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

Open in App

Solution

The correct option is **A** √6√2 ε0meρ

Given,

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.

Given,

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.

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program