Question

A particle that carries a charge $-q$ is placed at rest in uniform electric field $10N/C$. It experiences a force and moves in a certain time t, it is observed to acquire a velocity $10\overrightarrow{i}-10\overrightarrow{j}$ m/s. The given electric field intersects a surface of area $A$ $m^2$in the X -Z plane. Electric flux through surface is:

• A. $5\sqrt{2}AN{m}^2/C$
• B. $5AN{m}^2/C$
• C. $\sqrt{2}AN{m}^2/C$
• D. $2\sqrt{5}AN{m}^2/C$

Answer 1

The correct option is A $5\sqrt{2}AN{m}^2/C$

$v=10\hat{i}-10\hat{j}$

As velocity is in x and y components
Electric field also has two components

Let $E=Ex\hat{i}+Ey\hat{j}$

Now force on $(-q)=E(-q)$

Force $F=-Exq\hat{i}+-Eyq\hat{j}$

Acceleration $a=(-\frac{Exq}{m})\hat{i}+(-\frac{EyV}{m})\hat{i}$

Now $v_x=10$ and $v_y=-10$

we know $v=u+at(givenu=0)$

$\therefore V=at$

$\Rightarrow v_x=(-\frac{Exq}{m})t--\left(1\right)and{v}_y=\left(\frac{-Eyq}{m}\right)t--\left(2\right)$

$Given|E|=10N/C$

Now from 1 and 2 we get,

$-1=\frac{-Exq}{m}\frac{m}{Eyq}$

$\Rightarrow Ex=Ey$

$10=\sqrt{E{x}^2+E{y}^2}$

$\Rightarrow Ex=5\sqrt{2}=Ey$

Now than $\sigma =\int E.dA$

As the area is in X-Z plane

Area =$A$$\hat{i}$

$\therefore Ea=Ey.A$

$\varphi =5\sqrt{2}A$