Question

Three charges are arranged as shown in Figure. Find the magnitude and direction of the electrostatic force on the charge at the origin.

Answer 1

Force between two charges on x-axis is repulsion force, so the force on charge at origin acts in negative x-direction
\(\displaystyle{F}_{{1}}={k}{\left({6}\cdot{10}^{{-{9}}}\right)}{\frac{{{5}\cdot{10}^{{-{9}}}}}{{{\left({0.03}\right)}^{{2}}}}}={8.98}\cdot{10}^{{{9}}}\cdot{\left({6}\cdot{10}^{{-{9}}}\right)}{\frac{{{5}\cdot{10}^{{-{9}}}}}{{{\left({0.03}\right)}^{{2}}}}}\)
\(\displaystyle={2993.33}\cdot{10}^{{-{9}}}\) N
Force between two charges on y-axis is attractive force, so the force on charge at origin acts in negative y-direction
\(\displaystyle{F}_{{2}}={k}{\left({3}\cdot{10}^{{-{9}}}\right)}{\frac{{{5}\cdot{10}^{{-{9}}}}}{{{\left({0.01}\right)}^{{2}}}}}={8.98}\cdot{10}^{{{9}}}\cdot{\left({3}\cdot{10}^{{-{9}}}\right)}{\frac{{{5}\cdot{10}^{{-{9}}}}}{{{\left({0.01}\right)}^{{2}}}}}\)
angle made by the net force with positive x-axis = 180 + taninverse \(\displaystyle{\frac{{{13470}\cdot{10}^{{-{9}}}}}{{{2993.33}\cdot{10}^{{-{9}}}}}}={180}+{77.47}={257.77}\) degrees
magnitude of force \(\displaystyle=\sqrt{{{13470}^{{2}}+{2993.33}^{{2}}}}\cdot{10}^{{-{9}}}={13798.5}\cdot{10}^{{-{9}}}\) N