Wize University Physics Textbook (Master) > Electric Potential and Potential Energy

Electric Potential of Continuous Charge Distributions

Electric Potential of Continuous Charge Distributions

Similarly to electric field, we sometimes need to find the electric potential of a continuous charge distribution rather than a set of discrete point charges. This requires the use of integrals.

Steps to finding the electric potential of continuous distributions:

Draw a diagram if one is not already drawn in the question
Define one small charge element, $dq$ , on the diagram
Write down the electric potential, $dV$ , at the point of interest based on your charge element. Re-write all variables in terms of your integration variable.
Integrate your expression for the total electric potential: $V=\int dV$ . The bounds of the integrals are defined by the shape of the continuous charge distribution.

Example: Electric Potential of a Rod (on-axis)

Find the potential at point P due to the uniformly charged rod in the configuration below.

Let's add our charge element $dq$ to the diagram and re-write some expressions in terms of our integration variable, $x$ :

We start by determining the potential contribution from an infinitesimal piece of the rod, which has thickness $dx$ , is $x$ away from the origin, and contributes $dV$ to the total potential.

\begin{aligned} V&=\frac{kq}{r} \\ dV &= \frac{kdq}{l+a-x} \\ V &= \int_o^l\frac{k(\lambda dx)}{l+a-x} \\ V &= k \lambda \int_o^l\frac{dx}{l+a-x} \\ \end{aligned}

We integrate by substitution, using $u=l+a-x$ ( $du=-dx$ ), which changes our limits of integration to $u=l+a-(l)=a$ (upper) and $u=l+a-(0)=l+a$ (lower):

V = k \lambda \int_{l+a}^a\frac{-du}{u} \\ V = k \lambda \int_{a}^{l+a}\frac{du}{u} \\ V = k \lambda \ln\frac{l+a}{a} \\

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Practice: Electric Potential of a Rod (off-axis)

Find an expression for the electric potential at point P in the figure below. The rod is uniformly charged.

Illustration for practice question: Electric Potential of a Rod (off-axis)

You might find this integral useful:

\int \frac{1}{\sqrt{a^2+x^2}}dx=\ln(x+\sqrt{x^2+a^2})+C

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Practice: Rod with Positive and Negative Charge

A rod of length $l$ has a uniform charge distribution, except the left-hand side is positively charged and the right-hand side is negatively charged. Find the electric potential at point P which is at a distance $a$ from the origin.

Illustration for practice question: Rod with Positive and Negative Charge

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Practice: Electric Potential at the Center of a Half-Ring

What is the field at point P in the figure below? The total charge on the charge distribution is Q and it has a uniform positive charge distribution. Comment on your result.

Illustration for practice question: Electric Potential at the Center of a Half-Ring

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Practice: Electric Potential of a Ring

What is the electric potential at point P in the figure below? The total charge on the ring is Q and it has a uniform positive charge distribution.

Illustration for practice question: Electric Potential of a Ring

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Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Practice: Electric Potential of a Disk

a) What is the electric potential at point P in the figure below, at some distance L away from a uniformly charged disc of surface charge density $\sigma$ ?

b) What is the potential at point P if the distance L is much greater than the radius of the ring R?

Illustration for practice question: Electric Potential of a Disk

Remember that for a ring of radius a and total charge Q (uniformly distributed), the potential contribution to the point P was:

V=\frac{kQ}{\sqrt{a^2+L^2}}

Hint: You will find the following Taylor approximation useful (valid only for very small $x$ ):

\sqrt{1+x}\approx1+\frac12 x

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Electric Potential of Continuous Charge Distributions

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