Electrostatic Potential and Capacitance

Physics · Class 12

Lesson 3 of 13 · 7 min

Potential due to a system of charges

NCERT §2.5

Two beads sit on a ruler 20 cm apart: +3 nC at the 0 mark and −1 nC at the 20 cm mark. Meera wants the points on the ruler where the potential is exactly zero.

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In short

Superposition holds for potential: V at a point is the sum of the potentials of each charge on its own, V = (1/4πε₀) Σ qᵢ/rᵢ.

Potential is a scalar, so the sum is plain algebra with signs; no directions or components are needed. This is why potential is often easier to find than field.

A continuous distribution is split into small elements, each treated as a point charge, and the contributions are added (integrated).

Uniformly charged spherical shell of charge q and radius R: outside, V = q/(4πε₀r), as if all the charge sat at the centre.

Inside the shell E = 0, so no work is done moving a test charge inside, and V stays at its surface value q/(4πε₀R) throughout.

Worked value: 3 × 10⁻⁸ C and −2 × 10⁻⁸ C placed 15 cm apart give V = 0 at two points on their line: 9 cm and 45 cm from the positive charge.

Points of zero potential are not points of zero field: at 9 cm between the two charges above, both fields point the same way.

Potential due to a system of charges | Electrostatic Potential and Capacitance | Lumi Learn