Work to separate two charges

Problem

Drill: work to separate two charges. Two point charges $q_1 = +4\;\mu\text{C}$ and $q_2 = +3\;\mu\text{C}$ are initially at a distance $r_1 = 0{,}2\;\text{m}$ apart. How much work is needed to bring them to $r_2 = 0{,}5\;\text{m}$, keeping both stationary? Comment on the sign: who supplies the energy, the external agent or the field?

Solution

Potential energy of the pair. The electrostatic potential energy of two point charges at distance $r$ is $U(r) = \frac{kq_1 q_2}{r}.$ The work of the external agent, with kinetic energy unchanged (charges stationary before and after), equals the change in potential energy: $W_\text{est} = U(r_2) - U(r_1) = kq_1 q_2\left(\frac{1}{r_2} - \frac{1}{r_1}\right).$

Numerical calculation. With $kq_1 q_2 = 9\cdot 10^9 \cdot 4\cdot 10^{-6}\cdot 3\cdot 10^{-6} = 0{,}108\;\text{J}\cdot\text{m}$: $W_\text{est} = 0{,}108\left(\frac{1}{0{,}5} - \frac{1}{0{,}2}\right) = 0{,}108\,(2 - 5) = 0{,}108\cdot(-3) \approx \ev{W_\text{est} \approx -0{,}32\;\text{J}}.$

Sign. The two charges have the same sign, so they repel each other: as they are moved apart the Coulomb force acts in the same direction as the displacement and does positive work. It is the field that supplies the energy, not the external agent. The external work comes out negative: the operator must, if anything, brake the charges to keep them still, absorbing energy. Consistently, the system’s potential energy decreases (from $0{,}54$ J to $0{,}22$ J) as the charges move apart.

Connections

Topics: Electric field and potential Concepts: Electrostatic potential energy · Coulomb’s law Skills: Conservation of energy Methods: Gradient method Objects: Charged sphere