Question

A charged Q of 2.5uC is in two dimensional electric field with horizontal and vertical components of 35.0 N/C and 52.0 N/C, respectively. Find the magnitude and direction of the electric force on charge Q.

Answer 1

Now we know that

$E_{net}=\sqrt{E_1+E_2}$

$$E_{net}=\sqrt{35^2+52^2}=\sqrt{1225+2704}=\sqrt{3929}=62.68N/C$$

Force

$F=qE$

$$F=2.5\times10^{-6}\times62.68=1.56\times10^{-4}N$$

Direction

$$\phi =tan^{-1}(\frac{52}{35})=tan^{-1}(1.48)=56°$$