Question

A long cylindrical volume contains a uniformly distributed charge of density ρ. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic of the particle is :

(A) \(\frac{\rho qR^2}{4\varepsilon_0}\) 

(B) \(\frac{\rho qR^2}{2\varepsilon_0}\) 

(C) \(\frac{q\rho}{4\varepsilon_0R^2}\) 

(D) \(\frac{4\varepsilon_0R^2}{q\rho}\)

Answer 1

Correct option is (A) \(\frac{\rho q R^2}{4\varepsilon_0}\)

The charge particle is revolving around the cylinder in the electric field E.

So, we can write

\(qE = \frac{mv^2}r\)

Here, r is the distance of the particle from the axis of the cylinder.

Now, E = \(\frac{\rho r}{2\varepsilon_0}\)

⇒ \(\frac{mv^2}r = \frac{q\rho r}{2\varepsilon_0}\)

As, r = R

⇒ \(\frac12{mv^2}= \frac{\rho qR^2}{4\varepsilon_0}\)

Answer 2

Correct option is (A) \(\frac{\rho qR^2}{4\varepsilon_0}\)