Answer
a) $E_x=-2.5\times 10^2\frac{V}{m}$
b) $E_x=-5.00\times 10^2\frac{V}{m}$
c) $\theta=243^{\circ}$
Work Step by Step
(a) We know that
$E_x=-\frac{\Delta V}{\Delta x}$
We plug in the known values to obtain:
$E_x=-\frac{10.0}{0.0400}$
$E_x=-2.5\times 10^2\frac{V}{m}$
(b) We know that
$E_y=-\frac{\Delta V}{\Delta y}$
We plug in the known values to obtain:
$E_x=-\frac{10.0}{0.0200}$
$E_x=-5.00\times 10^2\frac{V}{m}$
(c) $E=\sqrt{(-250)^2+(-500)^2}=559\frac{V}{m}$
Now $\theta=tan^{-1}\frac{E_y}{E_x}$
$\implies \theta=tan^{-1}\frac{-500}{-250}=63.4^{\circ}+180^{\circ}$
$\theta=243^{\circ}$
