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(q_1q_2)/(d^2)`

Where k is the Coulomb's constant:

`k=8.99*10^9N\cdot(m^2)/(C^2)`

Isolate one of the charges from the equation:

`\Rightarrow q_2=(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*10^9N\cdot(m^2)/(C^2)^{})(-4.0\mu C)} \\ =\frac{(0.10m)^2(14.4N)}{(8.99*10^9N\cdot(m^2)/(C^2)^{})(-4.0*10^(-6)C)} \\ =-4.0*10^(-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`