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

Electric Potential Energy (with integrals)

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Electric Potential Energy - Integral Definition


Electric fields are not always constant and uniform. Even the electric field from a point charge, one of the most common electric fields, is not uniform! In order to describe the electric potential energy in these fields, we need to use more advanced techniques.
  • Work is defined as the dot product between a force and a displacement. If the force is constant and the path is a straight line, then this amounts to a simple multiplication. Otherwise, we need to consider an integral to add up each of the infinitesimal work amounts along the path between the two positions:
W=∫r1r2F⃗⋅dr⃗\boxed{W=\int_{r_1}^{r_2}\vec F \cdot d\vec r }
  • In electric fields, we can use F⃗=qE⃗\vec F = q \vec E to re-write the above (assuming that the charge is constant and can be pulled out of the integral):
W=q∫r1r2E⃗⋅dr⃗W=q\int_{r_1}^{r_2}\vec E \cdot d\vec r

Wize Tip
Because the electric force is conservative, the work done does not depend on the path you choose. If you are calculating work with an integral, pick the path between the two points that results in the easiest integral to solve, even if this is not the path that the charge actually took.

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  • For example, let's re-examine the electric potential energy between two point charges. If we have just one point charge q1q_1 in isolation, then there is no potential energy yet, and the electric field is as we expect (it points radially outward):
∣E⃗∣=kq1r2|\vec E| = k \frac{q_1}{r^2}

  • If we bring another point charge q2q_2nearby and we want to find the electric potential energy, we have to consider how much work is done in moving the charge from "very far away" (infinity) to some point r12r_{12} near the existing charge.
  • Note that the path is chosen to be along a field line, so the dot product becomes a simple multiplication for the entire integration.
W=q∫E⃗⋅dr⃗W=(q2)∫∞r12(kq1r2)drW=kq1q2∫∞r121r2drW=kq1q2[−1r]r=∞r=r12W=kq1q2(−1r12−0)W=−kq1q2r12\begin{aligned} W&=q\int\vec E \cdot d\vec r \\ W&=(q_2)\int_{\infty}^{r_{12}} (k\frac{q_1}{r^2}) dr \\ W&=kq_1q_2\int_{\infty}^{r_{12}} \frac{1}{r^2} dr \\ W&=kq_1q_2[-\frac{1}{r}]_{r=\infty}^{r=r_{12}} \\ W&=kq_1q_2(-\frac{1}{r_{12}}-0) \\ W&=-\frac{kq_1q_2}{r_{12}} \\ \end{aligned}
  • Finally, we recover the equation that was given as a fact in the last section.
ΔU=−W=kq1q2r12\begin{aligned} \Delta U&=-W=\frac{kq_1q_2}{r_{12}} \\ \end{aligned}

Practice: Electric Potential Energy of a Variable Field


Consider a two-dimensional electric field E⃗(x,y)=(xy,12x2)\vec{E}(x,y)=(xy, \frac12x^2).

How much work is done by the electric field in moving a +1.0 C charge from point (1,1)(1,1) to (2,3)(2,3)?

Practice: Electric Potential Energy of a Dipole


Consider a charge QQ that is placed a distance aa from the center of an electric dipole, along the axis of the electric dipole moment, closer to the positive charge than the negative charge of the dipole.

What is the change in electric potential energy of the charge QQ as it moves further away from the dipole to a distance bb?

Assume that both the initial and final positions are very far from the dipole, so the electric field approximation E=k2pr3E=k\frac{2p}{r^3} for a dipole along its axis holds throughout the entire path (that is, the distances in the above diagram are not to scale).
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Practice: Electric Potential Energy of a Continuous Rod


Consider a rod of length LL and uniform positive charge density λ\lambda.

a) How much work is required to bring a positive charge QQ from infinity to a distance dd away from the end of the rod, along the axis of the rod?


b) How much work would be required to bring the charge to a distance dd away from the center of the rod, along the perpendicular bisector of the rod?


Hint: you may find the following integral useful.
∫dxx2+a2=ln⁡(x+x2+a2)\int\frac{dx}{\sqrt{x^2+a^2}}=\ln(x+\sqrt{x^2+a^2})