 Work (electrical)
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Work Electrical

Electrical work is the work done on a charged particle by an electric field. The equation for 'electrical' work is equivalent to that of 'mechanical' work:

$W=Q\int _{a}^{b}\mathbf {E} \cdot \,d\mathbf {r} =Q\int _{a}^{b}{\frac {\mathbf {F_{E}} }{Q}}\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F_{E}} \cdot \,d\mathbf {r}$ where

Q is the charge of the particle, q, the unit charge
E is the electric field, which at a location is the force at that location divided by a unit ('test') charge
FE is the Coulomb (electric) force
r is the displacement
$\cdot$ is the dot product

The electrical work per unit of charge, when moving a negligible test charge between two points, is defined as the voltage between those points. The work can be done by electrochemical devices (electrochemical cells) or different metals junctions generating an electromotive force.

## Overview

### Qualitative overview

Particles that are free to move, if positively charged, normally tend towards regions of lower voltage (net negative charge), while if negatively charged they tend to shift towards regions of higher voltage (net positive charge).

However, any movement of a positive charge into a region of higher voltage requires external work to be done against the field of the electric force, which is equal to the work that the electric field would do in moving that positive charge the same distance in the opposite direction. Similarly, it requires positive external work to transfer a negatively charged particle from a region of higher voltage to a region of lower voltage.

The electric force is a conservative force: work done by a static electric field is independent of the path taken by the charge. There is no change in the voltage (electric potential) around any closed path; when returning to the starting point in a closed path, the net of the external work done is zero. The same holds for electric fields.

This is the basis of Kirchhoff's voltage law, one of the most fundamental laws governing electrical and electronic circuits, according to which the voltage gains and the drops in any electrical circuit always sum to zero.

### Mathematical overview

Given a charged object in empty space, Q+. To move q+ (with the same charge) closer to Q+ (starting from infinity, where the potential energy=0, for convenience), positive work would be performed. Mathematically:

$-{\frac {\partial U}{\partial \mathbf {r} }}=\mathbf {F}$ In this case, U is the potential energy of q+. So, integrating and using Coulomb's Law for the force:

$U=-\int _{r_{0}}^{r}\mathbf {F} \cdot \,d\mathbf {r} =-\int _{r_{0}}^{r}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {q_{1}q_{2}}{\mathbf {r^{2}} }}\cdot \,d\mathbf {r} ={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}}}\left({\frac {1}{r_{0}}}-{\frac {1}{r}}\right)+c$ c is usually set to 0 and r(0) to infinity (making the 1/r(0) term=0) Now, use the relationship

$W=-\Delta U\!$ To show that in this case if we start at infinity and move the charge to r,

$W={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}}}{\frac {1}{r}}$ 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

$P={\frac {\partial W}{\partial t}}={\frac {\partial QV}{\partial t}}$ V is the voltage. Work is defined by:

$\delta W=\mathbf {F} \cdot \mathbf {v} \delta t,$ Therefore

${\frac {\partial W}{\partial t}}=\mathbf {F_{E}} \cdot \,\mathbf {v}$ 