Next: Problems
Up: Electric Fields and Potentials
Previous: Electric Forces and Fields
We now place a charge q at y1 which moves, by the electric force, to the point y2 . The work done by the electric force is given by
W = Fd = qE(y2 - y1). | (6) |
W - PE = - (PE2 - PE1), | (7) |
PE q V = q(V2 - V1). | (8) |
E = - = - . | (9) |
From Eq.(1.7) we see that the units of electric potential energy PE are the same as those of work, which are Joules (J.). The units of potential difference V are found from Eq.(1.8) to be J/C, which is given the special name Volt (V). Eq.(1.9) allows us to then infer that the units of electric field E , which we earlier found to be N/C, can also be expressed as V/m.
We can from Eqs.(1.6,1.7,1.8) come up
with expressions giving the electric potential energy PE and potential V
at any point. Such expressions involve an arbitrary constant C :
PE = - qEy + C - PE = qE(y2 - y1) | |||
V = - Ey + C - V = E(y2 - y1). | (10) |
Although these considerations have been for a constant electric field, the
defining relation of Eq.(1.7) for the electric potential energy and
that of Eq.(1.8) for the electric potential are general. Note,
however,
that Eq.(1.9) relating the electric field to the potential difference
and the particular expressions of Eq.(1.10) for the potential
energy and potential hold only for a constant electric field. If the
electric field is not constant, these latter expressions change.
The derivation
involves some advanced mathematical techniques due to the fact that the
electric force changes from point to point. We quote here the corresponding
results for point charges. For this we imagine a point charge q moving from
a point A to a point B in the presence of a second point charge Q ,
as in Fig. 1.3.
We assume A is a distance r1 from Q and B is a distance
r2 from Q .
The work done by the electric force can be found using Coulomb's
law (1.2), which can subsequently be related to a change
in potential energy PE through Eq.(1.7). The
potential difference V can then be introduced through
Eq.(1.8). One finds
PE = k + C, | (11) | ||
V = k + C. |
A convenient and popular choice for the constant C in this instance is C = 0 , for which V(r = ) = 0 . The nice aspect of the electric potential energy and potential is that in the presence of multiple charges one simply adds algebraically the corresponding expressions for the individual charges to find the net result.
www-admin@theory.uwinnipeg.ca