Question

Protons are projected with an initial speed, $u={10}^4\text{m/s}$ into a region where a uniform electric field $\overrightarrow{E}=-720\hat{j}\text{N/C}$ is present, as shown in figure. The protons are to hit a target that lies at a horizontal distance of $1.3\text{mm}$ from the point where the protons are launched. Find the total time of flight for $\theta ={37}^{\circ }$.

• A. $0.7\times {10}^{-7}\text{s}$
• B. $1.7\times {10}^{-7}\text{s}$
• C. $2.7\times {10}^{-7}\text{s}$
• D. $3.7\times {10}^{-7}\text{s}$

Answer 1

The correct option is B $1.7\times {10}^{-7}\text{s}$

Given:
$\overrightarrow{E}=-720\hat{j}\text{N/C};m=1.6\times {10}^{-27}\text{kg}$

Proton is a positive charge particles, so it experienced an addition force in downward direction due to present electric field.

For the given situation,

$g_{eff}=g+\frac{qE}{m}$

$\Rightarrow g_{eff}=10+\frac{1.6\times {10}^{-19}\times 720}{1.6\times {10}^{-27}}=7.2\times {10}^{10}{\text{m/s}}^2$

From projectile motion, the total time of flight $T$ is

$T=\frac{2u\sin \theta }{g_{eff}}$

$\Rightarrow T=\frac{2\times {10}^4\times \sin {37}^{\circ }}{7.2\times {10}^{10}}$

$\Rightarrow T=1.7\times {10}^{-7}\text{s}$

Hence, option $\left(b\right)$ is the correct choice.