Total relation

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Electrical work is the work done on a charged particle by an electric field. The equation for 'electrical' work is, naturally, 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

F_E is the Coulomb (electric) force

r is the displacement

$\cdot$ is the dot product

Overview

Qualitative overview

Under normal circumstances, positively charged particles that are free to move will always tend to shift towards the direction of lower voltages (net negative charge). In contrast to this behavior of positive charges, particles that are negatively charged tend to shift towards regions that possess higher voltage (net positive charge).

However, it requires positive external work to move a positive charge into a region possessing a higher value of voltage. This is because external work is needed to be done against the field of the electric force; The work done here is equal to the work done by the electric field if it were to carry said 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 which is done by an 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 phenomenon holds good for electric fields as well.

The fact explained above forms the basis of one of the most fundamental laws governing electrical and electronic circuits, the Kirchhoff's voltage law. According to this principle, the voltage gains and the drops that occur around any electrical circuit loop is always equal to zero.

Mathematical overview

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. There is 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:

$U=-\int _{r_{0}}^{r}\mathbf {F} \cdot \,d\mathbf {r} =-\int _{r_{0}}^{r}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {q_{1}q_{2}}{\mathbf {r^{2}} }}\cdot \,d\mathbf {r} ={\frac {q_{1}q_{2}}{4\pi \varepsilon _{0}}}({\frac {1}{r_{0}}}-{\frac {1}{r}})+c$

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.

Note that although 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:

$W=(\mathbf {E} \cdot \,\mathbf {r} )=\mathbf {F_{E}} \cdot \,\mathbf {r}$

or 'force times distance' (times the cosine of the angle between them).

Electric Power

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church.

$P={\frac {\partial W}{\partial t}}={\frac {\partial QV}{\partial t}}$

V is the voltage. Work is defined by:

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

Therefore

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