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Wize University Physics Textbook (Master) > Electric Potential and Potential Energy

Conservation of Energy in Electricity

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Conservation of Energy


It should not be surprising that we can apply conservation of energy to problems in electrostatics. This will allows us to further link concepts in electricity with concepts from mechanics.
  • Just like in mechanics, the total mechanical energy of a moving charge in an external electric field is conserved when moving between two points. If potential energy increases, then kinetic energy decreases by the same amount, and vice-versa.
ΔU=−ΔKUi+Ki=Uf+KfUi+12mvi2=Uf+12mvf2\begin{aligned} \Delta U &= -\Delta K \\ U_i+K_i&=U_f+K_f \\ U_i+\frac12mv_i^2 &= U_f + \frac12 mv_f^2 \\ \end{aligned}
  • We have already seen that work is defined as the change in potential energy: W=−ΔU=−(Uf−Ui)W=-\Delta U=-(U_f-U_i).
  • Since mechanical energy is conserved, this means that work must also equal the change in kinetic energy, W=ΔKW=\Delta K (assuming there are no non-conservative forces present, such as friction or air resistance).

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Example: Releasing Two Repulsive Charges


Two point charges are held 20 cm apart and each carries 3.0 µc of positive charge. At the same time, they are released from rest, and they begin to move straight apart from each other. What is the velocity of the charges when they are 60 cm apart? Consider the mass of the charges to be 0.3 grams. Give your answer in m/s.

We can use conservation of energy to compare the initial configuration to the final configuration.

Initial configuration: Charges are static and not moving. There is electric potential energy in the system, but no kinetic energy.

Final configuration (f): The charges are moving (they each have kinetic energy), and there is also still some electric potential energy in the system. By symmetry, the velocity of the two charges must be equal.

Applying conservation of energy, remember that there are two objects that each have kinetic energy, but only one potential energy term to represent the potential energy between the two charges.

Ui+Ki=Uf+KfU12i+K1i+K2i=U12f+K1f+K2fkqqri+0+0=kqqrf+12mv2+12mv2kqqri−kqqrf=mv2v=kqqm(1r1−1r2)v=(8.99×109N m2C2)(3.0×10−6C)(3.0×10−6C)(0.0003kg)(10.20m−10.60m)vf=29.82m/s\begin{aligned} U_i+K_i&=U_f+K_f\\ U_{12i}+K_{1i}+K_{2i}&=U_{12f}+K_{1f}+K_{2f}\\ \cfrac{kqq}{r_i}+0+0&=\cfrac{kqq}{r_f}+\cfrac{1}{2}mv^2+\cfrac{1}{2}mv^2 \\ \cfrac{kqq}{r_i}-\cfrac{kqq}{r_f}&=mv^2 \\ v&=\sqrt{\cfrac{kqq}{m}\bigg(\cfrac{1}{r_1}-\cfrac{1}{r_2}\bigg)}\\ v&=\sqrt{\cfrac{(8.99\times10^9 \frac{N~m^2}{C^2})(3.0\times10^{-6} C)(3.0\times10^{-6} C)}{(0.0003kg)}\bigg(\cfrac{1}{0.20m}-\cfrac{1}{0.60m}\bigg)}\\ v_f&=29.82m/s \\ \end{aligned}

Practice: Conservation of Electric Potential Energy


Three point charges are fixed at corners of an isosceles triangle, as shown below. If q=3.0 μC and d=1.0 cm,
a) What is the electric potential energy of the system?
b) If the top charge is released, what would be its velocity when is very far from the other two charges? Consider the mass of each charge to be 1.0 gram.