Question

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

Short answer

The kinetic energy of the charged particle after 't' seconds is \(K = \left[\left(E^2q^2t^2\right) / (2m)\right]\).

Step-by-step solution

Electromagnetic Force

The electromagnetic force acting on the charged particle in the electric field can be found using the formula: \( F = qE \) where F is the electromagnetic force, q is the charge of the particle, and E is the electric field strength.

Acceleration

To find the acceleration of the charged particle, use Newton's second law: \( F = ma \) where 'F' is the electromagnetic force and 'a' is the acceleration of the particle. Therefore, 'a' can be found as : \( a = \frac{F}{m} = \frac{qE}{m} \)

Velocity

The velocity 'v' of the charged particle after 't' seconds can be found using the formula: \( v = at \) where 'a' is the acceleration we found earlier, and 't' is the time. Put the found value of acceleration in this formula: \( v = \frac{qEt}{m} \)

Kinetic Energy

Finally, we can find the kinetic energy 'K' of the charged particle using the formula: \( K = \frac{1}{2} mv^2 \) Put the value of 'v' we found in this formula: \( K = \frac{1}{2} m \left(\frac{qEt}{m}\right)^2 = \frac{1}{2} \frac{q^2E^2t^2}{m} \) So, the kinetic energy of the charged particle after 't' seconds is: \( K = \left[\left(E^2q^2t^2\right) / (2m)\right] \) Hence, the correct option is (B).