Question

A particle of mass 2 g and charge 6 $\mu$C is accelerated from rest through a potential difference of 60 V. The speed acquired by the particle is:

• A. \(0.6 \, {ms}^{-1}\)
• B. \(1.2 \, {ms}^{-1}\)
• C. \(1.8 \, {ms}^{-1}\)
• D. \(0.3 \, {ms}^{-1}\)

Hint:

To find the final speed of a charged particle accelerated through a potential difference, use the energy conservation principle: the kinetic energy gained is equal to the potential energy lost.

Answer 1

The Correct Option is A

The kinetic energy acquired by the particle is equal to the work done on it by the electric field, which is given by the change in electric potential energy: \[ KE = qV \] where \(q = 6 \times 10^{-6}\) C (charge of the particle), and \(V = 60\) V (potential difference). Calculating the kinetic energy: \[ KE = 6 \times 10^{-6} \times 60 = 0.36 \times 10^{-3} \, {Joules} \] Using the kinetic energy formula \(KE = \frac{1}{2} m v^2\), where \(m = 0.002\) kg (mass of the particle), we solve for \(v\) (velocity): \[ 0.36 \times 10^{-3} = \frac{1}{2} \times 0.002 \times v^2 \quad \Rightarrow \quad v^2 = \frac{0.36 \times 10^{-3} \times 2}{0.002} \] \[ v = \sqrt{0.36} \approx 0.6 \, {ms}^{-1} \]