Question

an electron (of charge -e) revolves around a long wire with uniform charge density lambda in a circular path of radius r . its kinetic energy is given by

Answer 1

Given :

Uniform charge density on the long wire around which the electron revolves =$\lambda$

The radius of the circular path in which electron is revolving = 'r'

To Find :

What is the magnitude of kinetic energy of this electron ?

Solution :

Let the mass of $e^-$ be = m

And let the velocity of electron be = v

The electric field generated by the long wire with uniform charge density is :

=$\frac{2K\lambda}{r}$         (Here  K is the dielectric constant  )

Now the force acting on the electron due to this electric field is :

=$q \times E$

=$e \times \frac{2K\lambda}{r}$       ( e is the magnitude of  charge on $e^-$ )

Since the electron is moving in a circular path , so  :

$\frac{mv^2}{r}=\frac{2 k\lambda}{r}.e$

Or, $\frac{mv^2}{2}= k\lambda}{e$

So  , K.E of electron = $K\lambda e$

Hence , the kinetic energy of the electron is $K\lambda e$ .