Given a charged object in empty space, Q+. To move q+
closer to Q+ (starting from r_0 = \infty , where the
potential energy=0, for convenience), we would have to apply an external force against the
Coulomb field and positive work would be performed. Mathematically, using the definition of a
conservative force, we know that we can relate this force to a
potential energy gradient as: :\frac{\partial U}{\partial \mathbf{r}} = \mathbf{F}_{ext} Where U(r) is the
potential energy of q+ at a distance r from the source Q. So, integrating and using
Coulomb's law for the force: :U(r) = \Delta U = \int_{r_0}^{r} \mathbf{F}_{ext} \cdot \, d \mathbf{r}= \int_{r_0}^{r} \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{\mathbf{r^2}} \cdot \, d \mathbf{r}= - \frac{q_1q_2}{4\pi\varepsilon_0}\left(\frac{1}{r_0}- \frac{1}{r}\right) = \frac{q_1q_2}{4\pi\varepsilon_0} \frac{1}{r} Now, use the relationship : W = -\Delta U \! To show that the external work done to move a point charge q+ from infinity to a distance r is: :W_{ext} = \frac{q_1q_2}{4\pi\varepsilon_0}\frac{1}{r} This could have been obtained equally by using the definition of W and integrating F with respect to r, which will
prove the above relationship. In the example both charges are positive; this equation is applicable to any charge configuration (as the product of the charges will be either positive or negative according to their (dis)similarity). If one of the charges were to be negative in the earlier example, the work taken to wrench that charge away to infinity would be exactly the same as the work needed in the earlier example to push that charge back to that same position. This is easy to see mathematically, as reversing the boundaries of integration reverses the sign.
Uniform electric field Where the electric field is constant (i.e.
not a function of displacement, r), the work equation simplifies to: :W = Q (\mathbf{E} \cdot \, \mathbf{r})=\mathbf{F_E} \cdot \, \mathbf{r} or 'force times distance' (times the cosine of the angle between them). ==Electric power==