For rings and disks we added field vectors piece by piece and had to watch every component. Potential is kinder. Each piece $dq$ contributes $k\,dq/r$, a plain number with no direction, and numbers just add. This is often the easiest road to the field too: find $V$ first, then take its slope.
Key idea
With $V = 0$ far away,
$$V = \int \frac{k\,dq}{r}.$$
Ring, on its axis: $V = \dfrac{kQ}{\sqrt{z^2 + R^2}}$. Disk, on its axis: $V = \dfrac{\sigma}{2\varepsilon_0}\left(\sqrt{z^2 + R^2} - z\right)$. Rod, on its own line, $a$ beyond the end: $V = k\lambda\ln\dfrac{a + L}{a}$.
Why it works
Each piece $dq$ is a point charge, and a point charge's potential is $k\,dq/r$. Potentials of several charges add because their fields add: $-\int(\vec{E}_1 + \vec{E}_2)\cdot d\vec{\ell}$ splits into two integrals. Since each term is a number (positive for positive charge), there is no cancelling by direction and no need for components.
Worked examples
Worked example: A ring, at the center and on the axis
A ring of radius $5.0\ \text{cm}$ carries $4.0\ \text{nC}$. Find $V$ at its center and on its axis $12\ \text{cm}$ from the center.
- At the center every piece is $R$ away: $V = \dfrac{kQ}{R} = \dfrac{(8.99 \times 10^{9})(4.0 \times 10^{-9})}{0.050} = 719\ \text{V}$.
- On the axis every piece is $\sqrt{z^2 + R^2} = \sqrt{0.12^2 + 0.05^2} = 0.13\ \text{m}$ away: $V = \dfrac{kQ}{0.13} = 277\ \text{V}$.
Answer: $719\ \text{V}$ and $277\ \text{V}$
Worked example: A rod, on its own line
A rod $0.30\ \text{m}$ long carries $6.0\ \text{nC}$. Find $V$ at a point on its line, $0.10\ \text{m}$ beyond one end.
- $\lambda = 6.0\ \text{nC}/0.30\ \text{m} = 20\ \text{nC/m}$. A piece $dx$ at distance $x$ adds $k\lambda\,dx/x$.
- $$V = \int_{0.10}^{0.40}\frac{k\lambda\,dx}{x} = k\lambda\ln\frac{0.40}{0.10} = (8.99 \times 10^{9})(20 \times 10^{-9})\ln 4$$
- $V = 249\ \text{V}$.
Answer: $V \approx 249\ \text{V}$
Worked example: A disk, from rings
A disk of radius $10\ \text{cm}$ carries $10\ \text{nC}$ spread evenly. Find $V$ on its axis $5.0\ \text{cm}$ from the center.
- A ring of radius $r$, width $dr$ holds $dq = \sigma\,2\pi r\,dr$ and adds $\dfrac{k\,dq}{\sqrt{z^2 + r^2}}$.
- $$V = k\sigma\pi\int_0^R\frac{2r\,dr}{\sqrt{z^2 + r^2}} = 2\pi k\sigma\left(\sqrt{z^2 + R^2} - z\right) = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{z^2 + R^2} - z\right)$$ (substitute $u = z^2 + r^2$).
- $\sigma = 3.18 \times 10^{-7}\ \text{C/m²}$ and $\sqrt{0.050^2 + 0.10^2} - 0.050 = 0.0618\ \text{m}$, so $V = \dfrac{3.18 \times 10^{-7}}{2(8.85 \times 10^{-12})}(0.0618) = 1110\ \text{V}$.
Answer: $V \approx 1110\ \text{V}$
Which road to take
- Charge with strong symmetry (sphere, long line, big sheet): find $\vec{E}$ with Gauss's law, then integrate it to get $V$.
- Charge without that symmetry (ring, disk, finite rod): add $k\,dq/r$ directly. It is a single ordinary integral, with no angles.
- Need the field too? Find $V$ along a line, then differentiate: the field is minus the slope of the potential.
Common mistake
Splitting potential into components. Writing $dV\cos\theta$, or cancelling the potentials of pieces on opposite sides, treats $V$ like a vector. It is not: a ring's center has $V = kQ/R$ even though the field there is zero, and that stays true if the charge is bunched on one side. Check: does every term in your sum have the sign of its own charge, with no angles in it? Then you are adding potentials correctly.