Electric Work - Yousef's Notes

When an electric force $\vec{\mathbf{F}}$ acts on a particle moving from a to b, the work done by the force is a line integral $$ W_{a \rightarrow b}=\int_{a}^{b} \overrightarrow{\mathbf{F}} \cdot {d \vec{l}}=\int_{a}^{b} F \cos \phi d l $$

where $d \vec{l}$ is an infinitesimal displacement along the particle’s path and $\phi$ is the angle between $\vec{\mathbf{F}}$ and $d\vec{\mathbf{l}}$.

(Recall that electric power is then $P=\frac{dW}{dt}$)

When a charged particle moves in an electric field, the field exerts a force that can do work on the particle.This work can always be expressed in terms of electric potential energy (see next section).

Combining the general expression with the definition of the potential,

$$ W_{a \rightarrow b}=\int_{a}^{b} \overrightarrow{\mathbf{F}} \cdot {d \vec{l}}=\int_{a}^{b} F \cos \phi d l=-q\int_{a}^{b} \nabla V(\mathbf{r}) \cdot {d \vec{l}}=U_{a}-U_{b}=-\left(U_{b}-U_{a}\right)=-\Delta U $$

Conclusion: electric force is conservative: it only depends on the difference between the initial and final points.

#Special Cases

#Work in a uniform electric field

$$ W_{a \rightarrow b}=F d=q_{0} E d $$

#Work in spherical field

$$ W_{a \rightarrow b}=\int_{r_{a}}^{r_{b}} F_{r} d r=\int_{r_{a}}^{r_{b}} \frac{1}{4 \pi \epsilon_{0}} \frac{q q_{0}}{r^{2}} d r=\frac{q q_{0}}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{r_{b}}\right) $$