A body having specific charge 8 μC/g is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V/m is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be _______________ s.

Solution

Physics sir · 4 steps

Find the acceleration from qE/m, time the trip to the wall, then double it for the return trip.

1. Convert specific charge. Specific charge is q/m. Here 8 μC/g = (8 × 10⁻⁶ C)/(10⁻³ kg) = 8 × 10⁻³ C kg⁻¹.
2. Find the acceleration. The only horizontal force is qE, so a = q/mE = 8 × 10⁻³ × 100 = 0.8 m s⁻², directed towards the wall.

   t = √(2d/a) = √((2 × 0.1)/0.8) = √0.25 = 0.5 s

3. Return trip. The body starts from rest and reaches the wall in 0.5 s. The elastic collision reverses its velocity without changing its speed. The same force now slows it down, so it comes to rest at its starting point after another 0.5 s. Then the motion repeats.
4. So. Time period = 0.5 + 0.5 = 1 s.

The common slip: 0.5. Stopping at the time to reach the wall gives 0.5 s. The period also includes the return trip back to rest, which takes another 0.5 s.

The idea to keep. A body under constant force bouncing elastically off a wall is like a ball dropped on the floor. One period is the trip to the wall plus the identical trip back.

Answer from the NTA final answer key. Solution by PrepWiser.