Question

An electron enters the space between the plates of a charged capacitor as shown in the figure. The surface charge
density on the plate is sigma. Electric field Intensity in the space between the plates is E. The uniform magnetic field
also exists in that space perpendicular to the direction of E. The electron moves perpendicular to both Ē and B
without any change in direction. The time taken by the electron to travel a distance in the space is
​

Answer 1

Given that,

Surface charge density = σ

Electric field = E

Magnetic field = B

We need to calculate the velocity of electron

Using relation of electrostatic and magnetic force

$qvB=qE$

Where, q = charge of electron

B = magnetic field

E = electric field

v = velocity of electron

$v=\dfrac{E}{B}$....(I)

We know that,

$E=\dfrac{\sigma}{\epsilon_{0}}$

Put the value of E in equation (I)

$v=\dfrac{\sigma}{\epsilon_{0}B}$

We need to calculate the time taken by the electron to travel a distance in the space

Using formula of time

$t = \dfrac{l}{v}$

Put the value into the formula

$t=\dfrac{l}{\dfrac{\sigma}{\epsilon_{0}B}}$

$t=\dfrac{l\epsilon_{0}B}{\sigma}$

Hence, The time taken by the electron to travel a distance in the space is $\dfrac{l\epsilon_{0}B}{\sigma}$.