Potential of Continuous Charge
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
For spread-out charge, add up each piece’s potential: V = ∫k dq/r. Because potential is a scalar, there are no components to cancel. That often makes V easier to find than E.
1. No vectors needed
2. When things are and aren’t simple
If every piece is the same distance from the point (ring center, ring axis, semicircle center), V = kQ/r right away, even though E might be zero there (ring center).
If the distance varies, write r in terms of the position variable and integrate.
Inside a solid charged sphere, find E first with Gauss’s law, then integrate it inward from infinity (next lesson): that’s how 3kQ/(2R) at the center comes out.
3. Example, then try it
A ring of radius 0.30 m carries 6.0 nC. Find the potential on its axis 0.40 m from the center, and at the center.
Distance: √(0.40² + 0.30²) = 0.50 m. V = (9.0 × 10⁹)(6.0 × 10⁻⁹) ÷ 0.50 = 108 V.
At the center: 54 ÷ 0.30 = 180 V (while E there is zero).
A rod of length L = 0.20 m carries Q = 4.0 nC spread uniformly. Find the potential on its axis, d = 0.10 m beyond one end.
Set up: rod from x = 0 to L, point at x = L + d. dV = kλ dx / (L + d − x), with λ = Q/L.
Integrate: V = kλ[−ln(L + d − x)]0L = kλ ln[(L + d)/d] = (kQ/L) ln[(d + L)/d].
Numbers: kQ/L = 36 ÷ 0.20 = 180 V, and ln(0.30/0.10) = ln 3 ≈ 1.10. V ≈ 198 V.
Check: a point charge 0.20 m away (the rod’s center) would give 180 V. Close, a bit higher because of the near end.
Answer: V = (kQ/L) ln[(d + L)/d] ≈ 198 V.
4. Common mistakes
Why is it often easier to find the potential of a charge distribution than its electric field?
V adds as plain numbers. For E, you must split each piece’s field into components and integrate each one.
Ready to practice?
5 questions on Potential of Continuous Charge, each with a hint and an explanation for every wrong answer.