Question

In a Cartesian plane, 4 charges are placed as follows: a 5C charge is
placed at (0, 3), -4C charge is placed at (5, -2), 1.5C charge is placed at
(0, 0) and a -6C charge is placed at (-4, -4). Determine the force felt by
the charge at the origin due to the presence of the other charges.

Answer 1

Explanations & Calculations

• Refer to the figure attached. The way those forces act on the middle charge is shown in it.

• Using Coulomb's law equation we can find the magnitude of each of the forces & then resolving each force in orthogonal directions the resultant force can be found.

$\qquad\qquad \begin{aligned} \small F_1&=\small (9\times10^9).\frac{1.5\times5}{3^2}=7.5\times10^9N\\ \small F_2&=\small (9\times10^9).\frac{1.5\times|-6|}{(4^2+4^2)}=2.5\times10^9N\\ \small F_3&=\small (9\times10^9).\frac{1.5\times|-4|}{(5^2+2^2)}=1.9\times10^9N\\\\ \small F_x&=\small F_2\cos\alpha- F_3\cos\beta\\ &=\small F_2\Big[\frac{4}{\sqrt{4^2+4^2}}\Big]-F_3\Big[\frac{5}{\sqrt{5^2+2^2}}\Big]\\ &=\small 3.7\times10^6N\\ \small F_y &=\small F_1+F_2\sin\alpha+F_3\sin\beta\\ &=\small F_1+F_2\Big[\frac{4}{\sqrt{4^2+4^2}}\Big]+F_3\Big[\frac{2}{\sqrt{5^2+2^2}}\Big]\\ &=\small 10.0\times10^9N\\\\ \small F_R&=\small \sqrt{F_x^2+F_y^2}=1.0\times10^{10}N\\ \small \tan\theta&=\small \frac{F_y}{F_x}\\ \small \theta&=\small 89.98^0[S\,of\,W] \end{aligned}$