Answer
Magnitude = $1.73\times10^{-8}$ [N]
Direction = $32.49^{\circ}$ North of East
Work Step by Step
Calculate the magnitude of the force.
$F = \frac{kq^{2}}{(1.5x10^{-10})^{2}}$
Since both forces are of equal radius,
$F_{1} = F_{2}$
Substitute known variables.
$F = \frac{(9x10^{9})(1.6x10^{-19})^{2}}{(1.5x10^{-10})^{2}}$
F = $1.024\times10^{-8} [N]$
The magnitude of q1 is decomposed q1 into x and y components:
X component = $cos65\times1.024\times10^{-8} [N]$
= $4.328x10^{-9}$ [N]
Y component = $sin65\times1.024\times10^{-8} [N]$
= $9.28x10^{-9}$ [N]
For q2, since it is in a straight x-axis, there is only an x component.
$F = \frac{(9x10^{9})(1.6x10^{-19})^{2}}{(1.5x10^{-10})^{2}}$
F = $1.024\times10^{-8} [N]$
Add the x components of both together.
$F_{xnet} = 1.024\times10^{-8} [N]$ + $4.328x10^{-9} [N]$
$F_{xnet} = 1.4568\times10^{-8}$
Since q1 is the only y component, (q2's y component is 0), it remains the same
To get the magnitude we use the Pythagorean theorem:
$\sqrt(1.4568\times10^{-8})^{2}+(9.28x10^{-9})^{2}
=1.73\times10^{-8} [N]$
To get the direction we use the inverse tangent function:
$tan_{-1}(\frac{9.28x10^{-9}}{1.4568\times10^{-8}})$
= $32.49^{\circ}$ North of East
