Question

An infinite dielectric sheet having surface charge
density (+0) has a hole of radius Rin it. An electron
(e) is released from rest on the axis of the hole at a
distance 3R from its centre. Then choose the
correct option(s). (mass of electron = m, charge on
electron = e)
The electron crosses the centre with speed
GeR
Vmeo
Motion of electron is SHM
Motion of electron is periodic but not SHM
The electron crosses the centre with the
20eR
speed
meo​

Answer 1

Given that,

Surface charge density = σ₀

Radius = R

Charge of electron = e

Distance from center = √3 R

We need to calculate the electric field

Using force equations electric field due to an infinite dielectric sheet

$E_{1}=\dfrac{\sigma_{0}}{2\epsilon_{0}}$.....(I)

Electric field at the axis of a hole of radius R is

$E_{2}=\dfrac{\sigma_{0}}{2\epsilon_{0}}(1-\dfrac{x}{\sqrt{x^2+R^2}})$..(II)

We need to calculate the resultant electric field

Using formula of resultant electric field

$E=E_{1}-E_{2}$

Put the value into the formula

$E=\dfrac{\sigma_{0}}{2\epsilon_{0}}-\dfrac{\sigma}{2\epsilon_{0}}(1-\dfrac{x}{\sqrt{x^2+R^2}})$

$E=\dfrac{\sigma_{0}}{2\epsilon_{0}}\dfrac{x}{\sqrt{x^2+R^2}}$

We need to calculate the speed of electron

Using formula of force on an electron

$F=-eE$

$F=-\dfrac{\sigma_{0}}{2\epsilon_{0}}\dfrac{ex}{\sqrt{x^2+R^2}}$

$mv\dfrac{dv}{dx}=-\dfrac{\sigma_{0}}{2\epsilon_{0}}\dfrac{ex}{\sqrt{x^2+R^2}}$

On integrating

$m\int_{0}^{v}{v\ dv}=-\dfrac{\sigma_{0}e}{2\epsilon_{0}}\int_{\sqrt{3}R}^{0}{\dfrac{x}{\sqrt{x^2+R^2}}}dx$

$\dfrac{mv^2}{2}=-\dfrac{\sigma_{0}e}{2\epsilon_{0}}(\sqrt{x^2+R^2})_{0}^{\sqrt{3}R}$

$v=\sqrt{\dfrac{\sigma_{0}\ eR}{m\epsilon_{0}}}$

Hence, The electron crossed the center with the speed is $\sqrt{\dfrac{\sigma_{0}\ eR}{m\epsilon_{0}}}$

(4) is correct option.