JEE Main 2021 — Electrostatics Question with Solution

Given,

q = 8$$\mu$$C/g = 8 $$\times$$ 10^$$-$$6 C/g = 8 $$\times$$ 10^$$-$$3 C/kg

s = 10 cm = 0.1 m $$\Rightarrow$$ E = 100 V/m

We know that, acceleration, a = $$\frac{force(F)}{mass(m)}$$

$$\Rightarrow$$ a = $$\frac{qE}{m}$$ [$$\because$$ F = qE]

= $$\frac{8\times {10}^{-6}\times 100}{{10}^{-3}}$$ = 0.8 ms^$$-$$2

As per question, when electric field is switched on, the body strikes to the wall and then returns back.

For one oscillation,

s = ut + $$\frac{1}{2}$$ at²

$$\Rightarrow$$ 0.1 = $$\frac{1}{2}$$ $$\times$$ 0.8 t² [$$\because$$ u = 0]

$$\Rightarrow$$ 0.2 = 0.8 t²

$$\Rightarrow$$ $$\frac{2}{8}$$ = t² $$\Rightarrow$$ t² = $$\frac{1}{4}$$ $$\Rightarrow$$ t = $$\frac{1}{2}$$

$$\therefore$$ Time period = 2 $$\times$$ $$\frac{1}{2}$$ = 1 s

Therefore, if the collision of the body is perfectly elastic, the time period of motion will be 1s.