Question

A ball of charge to mass ratio of$8\mathrm{\mu C}/\mathrm{g}$ is placed at a distance of $10\mathrm{cm}$ from the ball. An electric field of $100\mathrm{N}/\mathrm{m}$ is switched on in the direction of the wall. Find the time period of its oscillations. Assume all collisions are elastic.

• A. $1\sec$
• B. $2\sec$
• C. $3\sec$
• D. $4\sec$

Answer 1

The correct option is A

$1\sec$

Step 1: Given data and diagram

Charge to mass ratio of a ball$=\frac{\mathrm{q}}{\mathrm{m}}=8\mathrm{\mu C}/\mathrm{g}=\frac{8\times {10}^{-6}}{{10}^{-3}}\mathrm{C}/\mathrm{Kg}$

Distance between ball and wall$=\mathrm{S}=10\mathrm{cm}=0.1\mathrm{m}$

Electric field in the direction of wall, $\mathrm{E}=100\mathrm{N}/\mathrm{m}$

Let time period of oscillation$=\mathrm{T}$

Also let acceleration of a ball$=\mathrm{a}$

Step 2: Calculating the time period of oscillation

We know that the acceleration of a charged particle is given by:

$$\begin{gathered} \mathrm{a}=\frac{\mathrm{q}}{\mathrm{m}}\times \mathrm{E} \\ \Rightarrow \mathrm{a}=\frac{8\times {10}^{-6}}{{10}^{-3}}\times 100 \\ \Rightarrow \mathrm{a}=0.8{\mathrm{ms}}^{-2} \end{gathered}$$

Now when the electric field is applied, the ball will strike the wall and will come back again thus making one complete oscillation.

So the time taken by a ball to reach the wall can be calculated by using the equation of motion as follows:

$$\begin{gathered} \mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow 0.1=0+\frac{1}{2}\times 0.8\times {\mathrm{t}}^2(\because \mathrm{u}=0) \\ \Rightarrow 0.2=0.8\times {\mathrm{t}}^2 \\ \Rightarrow \mathrm{t}=0.5\mathrm{s} \end{gathered}$$

Therefore the time period of an oscillation is given by:

$$\begin{gathered} \mathrm{T}=2\times \mathrm{t} \\ \Rightarrow \mathrm{T}=2\times 0.5 \\ \Rightarrow \mathrm{T}=1\mathrm{s} \end{gathered}$$

Therefore the time period of an oscillation is equal to $1\sec$.

Hence, option(A) is the correct answer.