Question:
A body having specific charge 8 \(\mu\)C/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.
A body having specific charge 8 \(\mu\)C/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.
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Find the acceleration \(a = qE/m\), then the time to reach the wall from \(d=\tfrac12 at^2\). After the elastic bounce the body decelerates back to rest at the start, so the period is twice that time.
Updated On: Nov 4, 2026
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Correct Answer: 1
Solution and Explanation
Specific charge \(q/m = 8\ \mu\text{C/g} = 8\times10^{-6}\text{ C}/10^{-3}\text{ kg} = 8\times10^{-3}\) C/kg.
Acceleration \(a = \frac{qE}{m} = 8\times10^{-3}\times100 = 0.8\ \text{m/s}^2\).
Distance to the wall \(d = 0.1\) m, starting from rest: \(d=\frac12 a t^2 \Rightarrow t=\sqrt{\frac{2\times0.1}{0.8}} = 0.5\) s.
Speed at the wall \(v = at = 0.4\) m/s. The elastic collision reverses the velocity with the same magnitude, while the force still points towards the wall, so the body decelerates at \(0.8\ \text{m/s}^2\) and stops after another \(0.5\) s, exactly at its starting point (it travels \(v^2/2a = 0.1\) m).
Then the motion repeats. Time period \(T = 0.5 + 0.5 = 1\) s.
Answer: 1
Acceleration \(a = \frac{qE}{m} = 8\times10^{-3}\times100 = 0.8\ \text{m/s}^2\).
Distance to the wall \(d = 0.1\) m, starting from rest: \(d=\frac12 a t^2 \Rightarrow t=\sqrt{\frac{2\times0.1}{0.8}} = 0.5\) s.
Speed at the wall \(v = at = 0.4\) m/s. The elastic collision reverses the velocity with the same magnitude, while the force still points towards the wall, so the body decelerates at \(0.8\ \text{m/s}^2\) and stops after another \(0.5\) s, exactly at its starting point (it travels \(v^2/2a = 0.1\) m).
Then the motion repeats. Time period \(T = 0.5 + 0.5 = 1\) s.
Answer: 1
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