D.2.12 (HL)—Two-charge potential energy
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Use the pair-energy expression
For two point charges or spherical conductors represented at their centres, keep the signs of q1 and q2 and use centre separation r. The zero reference is infinite separation.
E_p=k\frac{q_1q_2}{r}
Worked example — two negative conductors
Radii 2.5cm and 1.5cm, separated by a 1.7cm surface gap, give r=5.7×10−2m. For charges −4.7×10−8C and −6.3×10−8C, Ep=(8.99×109)q1q2/r=+4.7×10−4J. Positive energy matches repulsion between like charges.
Interpret the sign
Like charges give positive potential energy because work is required to bring them together. Unlike charges give negative potential energy because the electric field releases energy as they approach. Increasing separation moves the energy toward zero.
Use changes in energy
When a charge configuration changes, calculate the final minus initial potential energy. The external work and work done by the electric field have opposite signs under a quasistatic convention. Use the actual separation between charge centres, not the physical radius of either object.
Common trap
Do not use absolute values for q1q2 before deciding the sign of the energy, and do not put r² in the potential-energy formula; r² belongs to Coulomb force.
Questions determine a charge from a potential-energy difference or evaluate energy changes in a charge configuration.
Determine
Use Ep=kq1q2/r with signed charges and centre separation, then compare final and initial energy if work is requested.
Dropping the sign of q1q2 or using inverse-square dependence for potential energy.
Representative question
Determine the charge Q of the sphere.
8.99×109×Q×(5.0×10−21−1.0×10−11)=1.1×103Q=1.2×10−8≪C>
The HL extension is secure when you can connect electric energy, potential and field geometry.