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 energy of the particle is:

• A. \(\frac{ρqR^2}{4∈_0}\)
• B. \(\frac{ρqR^2}{2∈_0}\)
• C. \(\frac{qρ}{4∈_0R^2}\)
• D. \(\frac{4∈_0R^2}{qρ}\)

Answer 1

The Correct Option is A

To solve this problem, we need to find the kinetic energy of a charge \( q \) revolving around a cylindrical volume containing a uniformly distributed charge density \( \rho \). The key here is to understand the electrostatic forces at play and how they relate to the kinetic energy of the particle.

1. First, let's understand the electric field inside a uniformly charged cylinder. The electric field \( E \) inside a uniformly charged cylindrical volume of charge density \( \rho \) at a distance \( r \) from the axis is given by Gauss's Law:

E = \frac{\rho r}{2∈_0}

2. However, the particle is revolving around the cylindrical surface at \( r = R \). Thus, the electric field at this location is:

E = \frac{\rho R}{2∈_0}

3. The force \( F \) on the charge \( q \) due to the electric field is:

F = qE = q \cdot \frac{\rho R}{2∈_0}

4. This force acts as the centripetal force required to keep the charge \( q \) moving in a circular path with radius \( R \). The centripetal force \( F_c \) is also given by:

F_c = \frac{mv^2}{R}

5. Setting the electric force equal to the centripetal force, we have:

q \cdot \frac{\rho R}{2∈_0} = \frac{mv^2}{R}

6. Solving for \( v^2 \), we get:

v^2 = \frac{q \rho R^2}{2∈_0 m}

7. The kinetic energy \( K.E. \) of the particle is given by:

K.E. = \frac{1}{2}mv^2

8. Substitute \( v^2 \) from the previous equation:

K.E. = \frac{1}{2}m \cdot \frac{q \rho R^2}{2∈_0 m} = \frac{q \rho R^2}{4∈_0}

9. Thus, the kinetic energy of the charge particle is:

\(\frac{ρqR^2}{4∈_0}\)

Therefore, the correct answer is the first option: \(\frac{ρqR^2}{4∈_0}\).