KAHIBARO
Table of Contents
From Point Charges to Continuous Charge
Electric potential is often first introduced for a single point charge. Real objects, however, usually have charge spread out over a length, an area, or a volume. In that case, we find the total potential by adding the contributions from many tiny pieces of charge.
The key idea is superposition. If a charge distribution is divided into small elements $dq$, then each element creates a small contribution to the potential at a point, and the total potential is the sum, or in the continuous limit, the integral of all these contributions.
For a small charge element $dq$ at distance $r$ from the observation point, the potential contribution is
$$
dV = \frac{1}{4\pi \varepsilon_0}\frac{dq}{r}
$$
so the total potential is
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{dq}{r}
$$
Here, $r$ means the distance from the charge element $dq$ to the point where the potential is being calculated.
For a continuous charge distribution, the electric potential is found from
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{dq}{r}
$$
This formula is one of the most important results in this chapter.
Why Potential Is Often Easier Than Field
When calculating electric field from a continuous distribution, direction must be handled carefully because the field is a vector. Electric potential is a scalar, so only ordinary addition is needed. That often makes potential calculations simpler.
A positive charge element gives positive potential contribution. A negative charge element gives negative potential contribution. After finding $V$, the electric field can later be obtained from the relation between field and potential, which belongs to a separate topic.
Charge Elements in Different Geometries
To use the integral, we must write $dq$ in a form that matches how the charge is distributed.
If charge is spread along a line, we use the linear charge density $\lambda$:
$$
\lambda = \frac{dq}{dl}
\quad \Rightarrow \quad
dq = \lambda \, dl
$$
If charge is spread over a surface, we use the surface charge density $\sigma$:
$$
\sigma = \frac{dq}{dA}
\quad \Rightarrow \quad
dq = \sigma \, dA
$$
If charge is spread through a volume, we use the volume charge density $\rho$:
$$
\rho = \frac{dq}{d\tau}
\quad \Rightarrow \quad
dq = \rho \, d\tau
$$
Here $d\tau$ is a small volume element.
| Distribution type | Density symbol | Definition | Charge element |
|---|---|---|---|
| Line charge | $\lambda$ | $\lambda = dq/dl$ | $dq = \lambda\,dl$ |
| Surface charge | $\sigma$ | $\sigma = dq/dA$ | $dq = \sigma\,dA$ |
| Volume charge | $\rho$ | $\rho = dq/d\tau$ | $dq = \rho\,d\tau$ |
Always choose the correct form of $dq$ before integrating:
$$
dq = \lambda\,dl, \qquad dq = \sigma\,dA, \qquad dq = \rho\,d\tau
$$
General Integral Forms
Using the density expressions, the potential can be written in three common ways.
For a line distribution,
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{\lambda \, dl}{r}
$$
For a surface distribution,
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{\sigma \, dA}{r}
$$
For a volume distribution,
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{\rho \, d\tau}{r}
$$
These formulas apply whether the charge density is constant or varies with position.
Strategy for Solving Problems
A typical calculation follows a clear pattern. First, choose the observation point where you want the potential. Next, select a small charge element $dq$. Then write the distance $r$ from that element to the observation point. After that, express $dq$ using the appropriate charge density and integrate over the whole distribution.
Because potential is a scalar, symmetry is still useful, but not in the same way as for vectors. Symmetry often simplifies the distance expression or the limits of integration.
Example, Potential on the Axis of a Uniform Ring
Consider a ring of radius $R$ carrying total charge $Q$, uniformly distributed. We want the potential at a point on the axis of the ring, at distance $x$ from its center.
Every small element $dq$ on the ring is the same distance from the observation point:
$$
r = \sqrt{R^2 + x^2}
$$
Since $r$ is constant for all elements, it can be taken outside the integral:
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{dq}{\sqrt{R^2 + x^2}}
= \frac{1}{4\pi \varepsilon_0}\frac{1}{\sqrt{R^2 + x^2}}\int dq
$$
But $\int dq = Q$, so
$$
V = \frac{1}{4\pi \varepsilon_0}\frac{Q}{\sqrt{R^2 + x^2}}
$$
This is a very useful example because the geometry makes the integral simple.
Example, Potential Due to a Uniformly Charged Rod
Suppose a rod lies along the $x$ axis from $x=a$ to $x=b$, with uniform linear charge density $\lambda$. We want the potential at the origin.
A small piece of rod of length $dx$ has charge
$$
dq = \lambda \, dx
$$
Its distance from the origin is $r = x$. Therefore,
$$
V = \frac{1}{4\pi \varepsilon_0}\int_a^b \frac{\lambda \, dx}{x}
$$
So,
$$
V = \frac{\lambda}{4\pi \varepsilon_0}\int_a^b \frac{dx}{x}
= \frac{\lambda}{4\pi \varepsilon_0}\ln\left(\frac{b}{a}\right)
$$
This result shows that the potential depends logarithmically on the rod endpoints.
When integrating for potential, the denominator must be the distance from each charge element to the observation point, not just a coordinate chosen by habit.
Example, Potential at the Center of a Uniformly Charged Disk
Consider a disk of radius $R$ with uniform surface charge density $\sigma$. We want the potential at the center.
A convenient method is to divide the disk into thin rings. A ring of radius $r$ and thickness $dr$ has area
$$
dA = 2\pi r \, dr
$$
so its charge is
$$
dq = \sigma \, dA = \sigma 2\pi r \, dr
$$
At the center, every point of that ring is at distance $r$ from the center, so
$$
dV = \frac{1}{4\pi \varepsilon_0}\frac{dq}{r}
= \frac{1}{4\pi \varepsilon_0}\frac{\sigma 2\pi r\,dr}{r}
$$
which simplifies to
$$
dV = \frac{\sigma}{2\varepsilon_0}dr
$$
Integrating from $0$ to $R$ gives
$$
V = \frac{\sigma}{2\varepsilon_0}\int_0^R dr
= \frac{\sigma R}{2\varepsilon_0}
$$
This example shows how choosing the right charge element can make the integral simple.
Variable Charge Density
Not all charge distributions are uniform. Sometimes the charge density changes with position. Then the only difference is that $\lambda$, $\sigma$, or $\rho$ must remain inside the integral.
For example, if a rod has linear charge density $\lambda(x)$, then
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{\lambda(x)\,dx}{r}
$$
The procedure is the same, but the algebra may be more involved.
Choice of Reference
Potential is defined relative to a reference level. For many isolated charge distributions, it is convenient to choose
$$
V(\infty) = 0
$$
Then the integral formula above gives the potential directly. This choice works well for finite charge distributions.
For some infinite charge distributions, the potential at infinity may not be finite, so extra care is needed. In such cases, potential differences are often more meaningful than absolute potential.
For finite charge distributions, a common and useful choice is
$$
V(\infty) = 0
$$
For infinite distributions, this choice may fail because the integral can diverge.
Common Difficulties
A common mistake is to confuse the integration variable with the distance $r$. They are sometimes the same, but often they are not. Another common mistake is forgetting that different parts of the distribution may lie at different distances from the observation point.
It is also important to choose a charge element that matches the symmetry of the problem. For a disk, rings are often better than little rectangular patches. For a spherical shell, angular elements may be better than Cartesian volume pieces.
Physical Interpretation
The integral for continuous charge distributions says that each tiny piece of charge contributes to the total potential according to its amount of charge and its distance from the point of interest. Nearby charge elements contribute more strongly because of the factor $1/r$. The total potential is the accumulated effect of the whole distribution.
Because potential is a scalar, positive and negative charge contributions combine through ordinary algebraic addition. This makes potential a powerful tool for handling extended charge distributions.
Summary
For a continuous distribution of charge, the electric potential is found by summing the contributions from infinitesimal charge elements:
$$
V = \frac{1}{4\pi \varepsilon_0}\int \frac{dq}{r}
$$
The form of $dq$ depends on the geometry of the distribution:
$$
dq = \lambda\,dl, \qquad dq = \sigma\,dA, \qquad dq = \rho\,d\tau
$$
The main task in any problem is to describe the geometry correctly, write the distance from each charge element to the observation point, and integrate over the entire distribution.