Question

A proton is released at rest $10\text{cm}$ from a charged sheet (perpendicular to it) having uniform electric field around it of intensity $123\text{N/C}$ towards (perpendicular to) the sheet. It will strike the sheet after the time (approximately)
(Charge on proton is $1.6\times {10}^{-19}\text{C}$ and mass of proton is $1.6\times {10}^{-27}\text{kg}$)

• A. $4\mu \text{s}$
• B. $2\mu \text{s}$
• C. $2\sqrt{2}\mu \text{s}$
• D. $4\sqrt{2}\mu \text{s}$

Answer 1

The correct option is A $4\mu \text{s}$

Given:
Intensity of electric field $E=123\text{N/C};$
Distance of the particle perpendicular to plate is $s=10\text{cm}=10\times {10}^{-2}\text{m}$

As we know that

$F=qE$

$\Rightarrow F=1.6\times {10}^{-19}\times 123=1.968\times {10}^{-17}\text{N}$

From Newton's law of motion

$F=ma$

$\Rightarrow a=\frac{F}{m}$

$\Rightarrow a=\frac{1.968\times {10}^{-17}}{1.6\times {10}^{-27}}=1.23\times {10}^{10}{\text{m/s}}^2$

Further, from equation of motion

$s=ut+\frac{1}{2}a{t}^2$

from question, $u=0$

$\Rightarrow 10\times {10}^{-2}=0+\frac{1}{2}\times 1.23\times {10}^{10}\times t^2$

$\Rightarrow t=\sqrt{\frac{2\times 10\times {10}^{-2}}{1.23\times {10}^{10}}}\approx 4\times {10}^{-6}\text{s}$

$\Rightarrow t\approx 4\mu \text{s}$

Hence, option $\left(a\right)$ is the correct answer.