Question

A repulsive force of 14.4 N occurs between two charges that are 0.10 m apart. If one charge is −4.0 μC , what is the other charge?−4.0 μC−1.0 μC+2.0 μC+1.0 μC

Answer 1

Use the Coulomb's Law to find the other charge:

$F=k\cdot\frac{q_1q_2}{d^2}$

Where k is the Coulomb's constant:

$k=8.99\times10^9N\cdot\frac{m^2}{C^2}$

Isolate one of the charges from the equation:

$\Rightarrow q_2=\frac{d^2F}{kq_1}$

Substitute d=0.1m, F=14.4N, q_1=-4.0μC and the value of k:

$\begin{gathered} q_2=\frac{(0.10m)^2(14.4N)}{(8.99\times10^9N\cdot\frac{m^2}{C^2}^{})(-4.0\mu C)} \\ =\frac{(0.10m)^2(14.4N)}{(8.99\times10^9N\cdot\frac{m^2}{C^2}^{})(-4.0\times10^{-6}C)} \\ =-4.0\times10^{-6}C \\ =-4.0\mu C \end{gathered}$

Notice that since the force is repulsive, both chargest must have the same sign.

Therefore, the answer is:

$-4.0\mu C$