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

Electric Potential of Continuous Charge Distributions

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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. 

Wize Concept
The steps to find electric potential in this way are very similar to those for finding electric field, with the notable difference that electric potential is a scalar, therefore we don't need to worry about splitting any vectors into components. 

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, dqdqdq, on the diagram
Write down the electric potential, dVdVdV, 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=∫dVV=\int dVV=∫dV. The bounds of the integrals are defined by the shape of the continuous charge distribution.
Exam Tip
Similarly to continuous electric field problems, some exam problems actually ask you to write the integral, but not solve them. For example, in some multiple choice problems, you will have to pick an integral that is set up properly, rather than having to determine the electric potential itself.

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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 elementdqdqdq to the diagram and re-write some expressions in terms of our integration variable, xxx:


We start by determining the potential contribution from an infinitesimal piece of the rod, which has thickness dxdxdx, is xxx away from the origin, and contributes dVdVdVto the total potential.
V=kqrdV=kdql+a−xV=∫olk(λdx)l+a−xV=kλ∫oldxl+a−x\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}VdVVV​=rkq​=l+a−xkdq​=∫ol​l+a−xk(λdx)​=kλ∫ol​l+a−xdx​​
We integrate by substitution, using u=l+a−xu=l+a-xu=l+a−x (du=−dxdu=-dxdu=−dx), which changes our limits of integration to u=l+a−(l)=au=l+a-(l)=au=l+a−(l)=a (upper) and u=l+a−(0)=l+au=l+a-(0)=l+au=l+a−(0)=l+a (lower):
V=kλ∫l+aa−duuV=kλ∫al+aduuV=kλln⁡l+aaV = 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} \\V=kλ∫l+aa​u−du​V=kλ∫al+a​udu​V=kλlnal+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.

You might find this integral useful:
∫1a2+x2dx=ln⁡(x+x2+a2)+C\int \frac{1}{\sqrt{a^2+x^2}}dx=\ln(x+\sqrt{x^2+a^2})+C∫a2+x2​1​dx=ln(x+x2+a2​)+C

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Practice: Rod with Positive and Negative Charge

A rod of length lll 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 distanceaaafrom the origin.

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.

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.

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

Remember that for a ring of radius a and total charge Q (uniformly distributed), the potential contribution to the point P was:
V=kQa2+L2V=\frac{kQ}{\sqrt{a^2+L^2}}V=a2+L2​kQ​

Hint: You will find the following Taylor approximation useful (valid only for very small xxx):
1+x≈1+12x\sqrt{1+x}\approx1+\frac12 x1+x​≈1+21​x

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