Practice Group Work

GWA Practice: Electric Potential on the Axis of a Charged Ring

Setup. A thin ring of radius \(R\) lies in the \(xy\)-plane, centered at the origin, with total charge \(Q\) uniformly distributed. Point \(P\) is on the \(z\)-axis at height \(z\). Use symmetry freely.

Figures

Center dq dl R θ
Top view: \(dl=R\,d\theta\) is tangent to the ring at \(dq\); radius \(R\) points from the center to \(dq\).
-R +R P(0,0,z) r x z
Side view: \(r=\sqrt{R^2+z^2}\) is the same for every element.
Integrate \( \theta \) from \(0 \to 2\pi\)
Uniform ring: integrate over the full angle.
dq dq in-plane components cancel
Only the axial component survives by symmetry.

Questions

  1. Write an expression for a small charge element \(dq\) in terms of the linear charge density \(\lambda\) and arc length \(dl\).
    Show solution
    \(dq=\lambda\,dl,\quad \lambda=\dfrac{Q}{2\pi R},\quad dl=R\,d\theta.\)
  2. Write the contribution \(dV\) to the potential at \(P\) from \(dq\). What simplification arises from the geometry?
    Show solution
    \(dV=\dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r}\), with \(r=\sqrt{R^2+z^2}\) constant for all \(dq\). All contributions are identical and add directly.
  3. Set up the integral for \(V(P)\). State the limits of integration and why the integral is straightforward.
    Show solution
    \(V(P)=\dfrac{1}{4\pi\varepsilon_0}\int\dfrac{dq}{\sqrt{R^2+z^2}} =\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{\sqrt{R^2+z^2}}.\) Limits: \(\theta:0\to 2\pi\). Trivial because \(r\) is constant.
  4. Give \(V(P)\) in terms of \(Q\), \(R\), \(z\), and \(\varepsilon_0\).
    Show solution
    \[ V(P)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{\sqrt{R^2+z^2}} \]
  5. Use \(E_z=-\,dV/dz\) to find the on-axis field.
    Show solution
    \[ E_z=-\frac{d}{dz}\!\left(\frac{1}{4\pi\varepsilon_0}\frac{Q}{\sqrt{R^2+z^2}}\right) =\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}. \]
  6. Check limiting cases: (a) \(z\gg R\). (b) \(z=0\).
    Show solution
    (a) \(V\approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{z}\) (point-charge limit).
    (b) \(V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\) at the center.
  7. Bonus: Replace the full ring with a half-ring (uniformly charged). How do \(V(P)\) and \(E_z\) change?
    Show solution
    Potential adds as scalars: if total charge is fixed at \(Q\), then \(V(P)\) is unchanged. If \(\lambda\) is fixed, the half-ring has \(Q/2\), so \(V\) halves. Axial field \(E_z\) follows the same scaling. Off-axis, symmetry is broken so lateral components don’t cancel.