The Physics Problem Solver/Electromagnetism/Electric Charge and Coulomb's Law

The scalar form of Coulomb's law is an expression for the magnitude and sign of the electrostatic force between two idealized point charges, small in size compared to their separation. This force (F) acting simultaneously on point charges (q₁) and (q₂), is given by

$F=k_{\mathrm {e} }{\frac {q_{1}q_{2}}{r^{2}}}$

where r is the separation distance and k_e is a proportionality constant. A positive force implies it is repulsive, while a negative force implies it is attractive. The proportionality constant k_e, called the Coulomb constant (sometimes called the Coulomb force constant), is related to defined and can be calculated based on knowledge of empirical measurements of the speed of light:

$k_{e}={\frac {1}{4\pi \varepsilon _{0}}}={\frac {c^{2}\ \mu _{0}}{4\pi }}=8.987\times 10^{9}\ \mathrm {N\cdot m^{2}/C^{2}} .$

Exercises

Beginner

1) Two point charges are initially 9 cm apart and are then moved so they are 2 cm apart. If the initial force between them is F what is the new force in terms of the initial force?

Using $F=k_{\mathrm {e} }{\frac {q_{1}q_{2}}{r^{2}}}$

For the initial state, r = 9cm = 0.09m

$F_{i}=k_{\mathrm {e} }{\frac {q_{1}q_{2}}{0.09^{2}}}$

For the initial state, r = 2cm = 0.02m

$F_{f}=k_{\mathrm {e} }{\frac {q_{1}q_{2}}{0.02^{2}}}$

Multiplying by r² in both cases gives us:

$F_{i}*0.09^{2}=k_{\mathrm {e} }q_{1}q_{2}\,$

$F_{f}*0.02^{2}=k_{\mathrm {e} }q_{1}q_{2}\,$

Setting the two equations to be equal ( they are both equal to $k_{\mathrm {e} }q_{1}q_{2}\,$):

$F_{i}\times 0.09^{2}=F_{f}\times 0.02^{2}\,$

$F_{f}={\frac {0.09^{2}}{0.02^{2}}}\times F_{i}=20.25F_{i}\,$

In other words, the new force is roughly 20x as strong as the initial force.