Problem 25 - A Complete Resource Book in Physics for JEE Main

Question

A charged particle of mass \(m\) and charge \(q\) is released from rest in an electric field of constant magnitude \(E\). The kinetic energy of the particle after a time \(t\) is (A) \(\frac{E^{2} q^{2} t^{2}}{m}\) (B) \(\frac{2 E^{2} q^{2} t^{2}}{m}\) (C) \(\frac{E^{2} q^{2} t^{2}}{2 m}\) (D) \(\frac{4 E^{2} q^{2} t^{2}}{m}\)

Step-by-Step Solution

Answer

The short answer for the problem is: The kinetic energy of the charged particle after a time \(t\) is \(\frac{E^{2} q^{2} t^{2}}{2 m}\).

Step 1: Identify the force on the particle

In the presence of an electric field, the force experienced by a charged particle is given by \(F = qE\), where \(F\) is the force experienced by the particle, \(q\) is its charge, and \(E\) is the magnitude of the electric field.

Step 2: Determine the particle's acceleration

Using Newton's second law, the acceleration of the particle can be calculated as \(a = \frac{F}{m}\), where \(a\) is the acceleration and \(m\) is the mass of the particle. Given the force from step 1, we can find the acceleration: \(a = \frac{qE}{m}\)

Step 3: Find the velocity after time t

The particle is initially at rest, so its initial velocity is \(0\). We can find the velocity after time \(t\) using the formula \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, and \(a\) is the acceleration: \(v = 0 + at = (qE)(\frac{t}{m})\)

Step 4: Calculate the kinetic energy after time t

The kinetic energy of a particle can be determined using the formula \(K = \frac{1}{2}mv^2\), where \(K\) is the kinetic energy and \(v\) is the velocity of the particle. Using the velocity we calculated in step 3, we can determine the kinetic energy after a time \(t\): \(K = \frac{1}{2}m(qE\frac{t}{m})^2\) From this expression, we can simplify and compare it with the answer choices given: \(K = \frac{1}{2}m\frac{q^2E^2t^2}{m^2}\) \(K = \frac{q^2E^2t^2}{2m}\) This expression matches choice (C). So, the correct answer is (C) \(\frac{E^{2} q^{2} t^{2}}{2 m}\).