Answer
(a) $559V/m,243^{\circ}$
(b) $8.95mm$
Work Step by Step
(a) We can find the required magnitude and direction of the electric field as follows:
$E_x=\frac{\Delta V}{\Delta x}$
$E_x=\frac{10V}{4\times 10^{-2}m}=-250V/m$
and $E_y=\frac{\Delta V}{\Delta y}$
$E_y=\frac{-10V}{2\times 10^{-2}m}=-500V/m$
Now $E=\sqrt{E_x^2+E_y^2}$
$\implies E=\sqrt{(-250)^2+(-500)^2}=559V/m$
$\theta=tan^{-1}(\frac{-500}{-250})=63.43^{\circ}$
The required direction is $\theta =180^{\circ}+43.63^{\circ}=243^{\circ}$
(b) We can find the required distance as follows:
$\Delta S=\frac{\Delta V}{E}$
$\Delta S=\frac{5.0V}{559V/m}=8.95mm$
