2.4 Electric Potential Due to a Continuous Charge Distribution

From point charges to continuous distributions

For a continuous charge distribution, treat each infinitesimal piece dq as a point charge and use V = kdq/r, then integrate over the whole object: V = k∫dq/r. Choose dq based on geometry: dq = λdl (line), dq = σdA (surface), or dq = ρdV (volume). Electric potential is a scalar, so contributions simply add — no vector components needed, unlike the electric field integral.

Setting up the integral

Express r, the distance from dq to the point where you want V, in terms of a single variable (like x). Substitute dq and r into V = k∫dq/r, then integrate over the object's limits. Common examples: a uniformly charged rod (V = kQ/L·ln[...]) and a uniformly charged ring (V = kQ/√(R²+x²) along the axis).

Using V to find E

Once V(r) is known, you can find the electric field using E = -dV/dr (or the gradient in multiple dimensions). This is often easier than direct field integration since potential avoids vector components. For a ring's axial potential V = kQ/√(R²+x²), differentiating gives the familiar axial field expression.

Key terms

General formula for V from continuous charge

V = k∫dq/r, integrating contributions from each infinitesimal charge element dq.

dq for a line, surface, and volume charge

Line: dq = λdl; Surface: dq = σdA; Volume: dq = ρdV.

Why is finding V easier than finding E directly?

V is a scalar, so you just add contributions — no need to break into vector components like with E.

Potential on the axis of a charged ring (radius R, charge Q)

V = kQ/√(R² + x²), since every point on the ring is equidistant from the axial point.

Relating E to V

E = -dV/dr (or -∇V in 3D); differentiate the potential function to get the field.

Try an AP-style question

A ring of radius R has charge Q. What is the potential on its axis a distance x from its center?

1. kQ/x
2. kQx/(R² + x²)^(3/2)
3. kQ/√(R² + x²)
4. kQ/(R² + x²)