Lesson 1 of 13 · 7 min

Electrostatic potential

NCERT §2.1 to §2.3

Meera switches on the lab's Van de Graaff generator. Its metal dome, 15 cm in radius, collects +5 μC, and her hair starts to lift when she stands near it. Carrying a small charged bead towards that dome takes effort. How much, and does the route matter?

The story this chapter follows: Meera's lab afternoon

Meera spends an afternoon in the physics lab with three things: a Van de Graaff dome 15 cm in radius holding +5 μC, which sits at 300 kV; a pair of 10 cm × 10 cm demo plates 2 mm apart on a 100 V supply; and a box of 10 μF capacitors with a 12 V battery. Potential, equipotentials, conductors and every capacitor idea in this chapter turn up on that bench.

The lesson in notes

In short

The Coulomb force is conservative: the work it does between two points depends only on the end points, not on the path taken.

Potential energy difference: the work an external agent does to move a charge from R to P slowly (no acceleration) against the field. U_P − U_R = W_RP, and the field's own work is −W_RP.

Only differences of potential energy matter, so a constant can be added freely. The usual choice puts zero potential energy at infinity.

Electrostatic potential V at a point: the work done by an external agent, per unit positive test charge, to bring it from infinity to that point. SI unit volt; 1 V = 1 J C⁻¹.

The work to take a charge q from R to P is q(V_P − V_R). The test charge is taken very small so it does not disturb the charges producing the field.

Point charge Q at the origin: V(r) = Q/(4πε₀r). V is positive for Q > 0 and negative for Q < 0, and it falls as 1/r while the field falls as 1/r².

V depends only on distance from a point charge, so it is the same at every point of a sphere centred on the charge.

Worked value: 4 × 10⁻⁷ C gives V = 9 × 10⁹ × 4 × 10⁻⁷ / 0.09 = 4 × 10⁴ V at 9 cm; bringing 2 × 10⁻⁹ C there from infinity needs 8 × 10⁻⁵ J, whatever the path.