Question

A test charge of +2 μC is placed halfway between a charge of +6 μC and another of +4 μC separated by

10 cm.

(a) What is the magnitude of the force on the test charge?

(b) What is the direction of this force (away from or toward the +6 μC charge)?

Answer 1

The force acting on a test charge is a sum of forces due to every charge. Let's calculate them separately :

1. $F_{2; 6} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2}=\frac{9\cdot10^9 \cdot 2\cdot10^{-6}\cdot6\cdot10^{-6}}{5^2\cdot10^{-4}}=43.2 N$, both charges are positive so this force repulses the test charge from the $6 \mu C$ charge.
2. $F_{2;4} = \frac{1}{4\pi\epsilon_0} \frac{q_2 q_3}{r^2} = \frac{9\cdot10^9\cdot2\cdot10^{-6}\cdot4\cdot10^{-6}}{5^2 \cdot 10^{-4}}=28.8N$, both charges are positive so this force repulses the test charge from the $4 \mu C$ charge and thus is directed towards the $6\mu C$ charge.

Therefore the resulting force is $F=F_{2;6}-F_{2;4}=14.4N$ and is directed from the $6 \mu C$ charge.