Question

There is a uniform electric field of strength of ${10}^3{\mathrm{Vm}}^{-1}$ along the y−axis. A body of mass $1\mathrm{g}$ and charge ${10}^{-6}\mathrm{C}$ is projected into the field from the origin along the positive$x$−axis with a velocity of $10\mathrm{m}/\mathrm{s}$. Find its speed in $\mathrm{m}/\mathrm{s}$ after $10\sec$ is (neglect gravitation).

Answer 1

Step 1: Given parameters.

Velocity in $x$-direction, $V_x=10\mathrm{m}/\mathrm{s}$

Electric field, $E={10}^3\mathrm{V}/\mathrm{m}$

Charge, $q={10}^{-6}\mathrm{C}$

Mass, $\mathrm{m}=1\mathrm{g}$

Step 2: Calculate the speed of body after $10\sec$.

As velocity in the $x$-direction remains constant at $10\mathrm{m}/\mathrm{s}$

Now, the acceleration will be in the $y$-direction only,

$a=\frac{qE}{\mathrm{m}}...\left(1\right)$

where, $a$ is the acceleration of the body, $q$ is the charge of the body, $E$is the electric field and, $\mathrm{m}$ is the mass of the body.

Substitute the given values in equation (1).

$a=\frac{{10}^{-6}\times {10}^3}{1\times {10}^{-3}}$

$a=1\mathrm{m}/{\mathrm{s}}^2$

The velocity in the $y$-direction will be,

$V_y=a\times t$

$V_y=1\times 10$

$V_y=10\mathrm{m}/\mathrm{s}$

Therefore the final velocity of the body,

$V=10\hat{x}+10\hat{y}$

$\left|V\right|=10\surd 2\mathrm{m}/\mathrm{s}$

Hence, the speed of the body after $10\sec$ will be $10\surd 2\mathrm{m}/\mathrm{s}$.