Question

In a region of space an electric field line is in the shape of a semicircle of radius R. Magnitude of the field at all points is E. A particle of mass m having charge q is constrained to move along this field line. The particle is released from rest at A. Find the acceleration of the particle when it is at the midpoint of the path from A to B.

• A. $\frac{qE}{m}$
• B. $\frac{qE}{m}\sqrt{1 + \pi^2}$
• C. $\frac{qE}{m}\sqrt{1 + \frac{\pi^2}{4}}$
• D. $\frac{qE}{m}\sqrt{2 + \pi^2}$

Answer 1

Answer: (B)

$\frac{qE}{m}\sqrt{1 + \pi^2}$

Answer 2

The problem asks for the acceleration of a charged particle constrained to move along a semicircular electric field line. The particle is released from rest at point A and we need to find its acceleration at the midpoint of the path from A to B.

1. Identify the forces and accelerations:

   • The particle has mass m and charge q.

   • It moves along an electric field line, meaning the electric force F_e = qE is always tangential to the path.

   • Since the particle is constrained to move along a curved path, it experiences two components of acceleration:

     • Tangential acceleration (a_t): Due to the electric force F_e.
     • Centripetal acceleration (a_c): Due to the change in direction of velocity, directed towards the center of curvature.

2. Calculate the tangential acceleration (a_t):

   The electric force F_e = qE acts tangentially along the path. According to Newton's second law, F_e = m a_t. Therefore, the magnitude of the tangential acceleration is: $a_t = \frac{qE}{m}$ This tangential acceleration is constant in magnitude since q, E, and m are constant.

3. Calculate the speed (v) at the midpoint:

   The particle starts from rest at A. To find the centripetal acceleration, we need the speed v at the midpoint. We use the work-energy theorem. The path from A to B is a semicircle of radius R. The total arc length is πR. The midpoint of the path from A to B is halfway along this arc, so the arc length from A to the midpoint is s = \frac{1}{2} (\pi R) = \frac{\pi R}{2}. The work done by the electric force W is F_e \times s because the force is always tangential and its magnitude is constant. $W = qE \times \left(\frac{\pi R}{2}\right)$ According to the work-energy theorem, W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2. $qE \frac{\pi R}{2} = \frac{1}{2}mv^2$ $mv^2 = qE\pi R$ $v^2 = \frac{qE\pi R}{m}$

4. Calculate the centripetal acceleration (a_c) at the midpoint:

   The centripetal acceleration is given by a_c = \frac{v^2}{R}. Substitute the expression for v^2: $a_c = \frac{1}{R} \left(\frac{qE\pi R}{m}\right)$ $a_c = \frac{qE\pi}{m}$

5. Calculate the total acceleration at the midpoint:

   At the midpoint of the semicircle (the highest point), the tangent to the path is horizontal, and the radius is vertical. Therefore, the tangential acceleration a_t (which is along the tangent) and the centripetal acceleration a_c (which is along the radius, pointing towards the center) are perpendicular to each other. The magnitude of the total acceleration a is the vector sum of a_t and a_c: $a = \sqrt{a_t^2 + a_c^2}$ Substitute the values of a_t and a_c: $a = \sqrt{\left(\frac{qE}{m}\right)^2 + \left(\frac{qE\pi}{m}\right)^2}$ $a = \sqrt{\left(\frac{qE}{m}\right)^2 (1 + \pi^2)}$ $a = \frac{qE}{m}\sqrt{1 + \pi^2}$