Question

A spherical cavity of radius R/2 is made inside a fixed uniformly charged non-conducting sphere of radius R and volume charge density ρ as shown. A particle of mass m and charge –q is released from rest from centre of cavity. Time taken by it to hit the surface of cavity is

Answer 1

Answer:

root 6E.m

--------

pq

Explanation:

this is the answer of the question

Answer 2

Answer:

Time taken to reach center/surface of the cavity is = $\sqrt \frac{3 m \epsilon_o}{\rho q R}$

Explanation:

Electric field at center of cavity = Field due to complete sphere - Field due to cavity part = $E_{sphere}-E_{cavity}$ = $\frac{KQ^2}{r^2}-\frac{KQ^2}{r^2} = \frac{\rho \frac{4\pi r^3}{3} }{4\pi \epsilon_o (r)^2 } = \frac{\rho r}{3\epsilon_o}$

Force due to this field = $F$ = $q.E = \frac{q\rho r}{3\epsilon_o}$

acceleration produced = $a = \frac{F}{m}= \frac{q\rho r}{3m\epsilon_o}$

we know, $a = \frac{d^2 x}{d^2t} = \frac{vdv}{dx}$

$\frac{vdv}{dr}=\frac{q\rho R}{6m\epsilon_o}$

$vdv= \frac{q\rho r}{3m\epsilon_o}dr$

Integrating both sides,

$\int \limits^v_0 vdv= \int \limits^{\frac{R}{2}}_0 \frac{q\rho r}{3m\epsilon_o}dr$

$\implies \frac{v^2}{2}= \frac{\rho q }{3m\epsilon_o} \frac{1}{2}\frac{r^2}{2}$

$\implies v= \sqrt {\frac{\rho q r^2}{12m\epsilon_o} }$

Also we know, $v= \frac{dx}{dt} =\frac{dr}{dt}$

$\implies \sqrt {\frac{\rho q r^2}{12m\epsilon_o} } = \frac{dr}{dt}$

$\implies dt=\sqrt \frac{12 m \epsilon_o}{\rho q R^2}dr$

Integrating both sides,

$\implies \int dt=\int \sqrt\frac{12 m \epsilon_o}{\rho q R^2}dr$

$\implies t=\sqrt \frac{12 m \epsilon_o}{\rho q R^2}(\frac{R}{2} )$

$\implies t=\sqrt \frac{3 m \epsilon_o}{\rho q R}$

[Note: We can not use newton's equation of motion to find time beacause electric field is not constant and acceleration is also not constant.]