Problem: The potential gradient dV/dx determines the force on a given charge. This can be understood through the example of the following problem:
If a potential gradient of 11 volts/meter exists between two points over a displacement of .4500 meters,
What is the voltage between the points, and how much work will be done on a charge of 56 Coulombs as it moves between these points?
From the work done and the displacement, determine the average force on the charge.
Repeat these calculations for the same potential gradient with a displacement of 3.9 meters.
From potential gradient dV/dx and displacement ds, find an expression for the work done on a charge q as it moves through this displacement, and use this expression to find an expression for the force on the charge. Give the resulting force for a charge of 51 Coulombs within a potential gradient of 44000 volts/meter.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution
Solution:
A potential gradient of 11 volts/meter over a displacement of .4500 meters implies a potential difference of ( 11 volts/meter)( .4500 meters) = 4.95 volts.
The work done on a charge of 56 Coulombs will therefore be ( 4.95 J/C)( 56 Coulombs) = 277.2 J.
This work is the product of the average force on the charge and the .4500 meter displacement. That is, ( .4500 m)(ave force) = 277.2 J. It follows that the average force is 277.2 J / ( .4500 m) = 616.0 Newtons.
If the displacement is 3.9 meters the potential difference is ( 11 volts/meter)( 3.9 meters) = 42.9 volts and the work done is 56 Coulombs ( 42.9 volts) = 2402 J, and the average force is
2402 J / ( 3.9 m ) = 615.8 Newtons.
Thus the force is the same as before. The work done is increased by the same proportion as the displacement, so the force, which is the work per unit displacement, is the same as before.
So the work done is the product of the charge, the potential gradient and the displacement, and also the product of the force and the displacement. That is,
W = q (dV/dx) ds and W = Fave ds.
Therefore
q (dV/dx) ds = Fave ds
and
Fave = q (dV/dx).
This confirms that the potential gradient dV/dx determines the force on a given charge in the circuit.
Generalized Response: A potential gradient dV / dx over a displacement ds results in potential difference (dV / dx) ds. The work done to move charge q through this potential difference is W = charge * pot diff = q (dV / dx) ds, which is equal to Fave ds, where Fave is the average force exerted on the charge. Thus Fave ds = q (dV/dx) ds, and Fave = q (dV/dx).
.
.
.
.
.
.
.
.
.
.
Figure description: The figure below shows a charge q moving through a displacement `ds in the direction of a potential gradient dV / dx. The potential difference will be the product of the potential gradient and the displacement: `dV = dV / dx * `ds. The work done on the charge is therefore `dW = q `dV; this work is also the product `dW = Fave * `ds of the average force on the charge and the displacement. Setting these two expressions for work `dW equal we arrive at the conclusion that Fave = q dV/dx = q E.
E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.
