22.5 Voltage for nonuniform fields by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
The calculus-savvy reader will have no difficulty generalizing the field-voltage relationship to the case of a varying field. The potential energy associated with a varying force is
ΔPE=-∫Fdx, [one dimension]
so for electric fields we divide by q to find
ΔV=-∫Edx, [one dimension]
Applying the fundamental theorem of calculus yields
E=-dVdx [one dimension]
⇒ What is the voltage associated with a point charge?
⇒ As derived previously in self-check A on page 625, the field is
|E|=kQr2
The difference in voltage between two points on the same radius line is
ΔV=∫dV
=-∫Exdx
In the general discussion above, x was just a generic name for distance traveled along the line from one point to the other, so in this case x really means r.
ΔV=∫r2r1Erdx
=∫r2r1kQr2dr
=kQr]r2r1
ΔV=kQr2-kQr1.
The standard convention is to use r1 = ∞ as a reference point, so that the voltage at any distance r from the charge is
The interpretation is that if you bring a positive test charge closer to a positive charge, its electrical energy is increased; if it was released, it would spring away, releasing this as kinetic energy.
self-check:
Show that you can recover the expression for the field of a point charge by evaluating the derivative Ex=-dVdx.
(answer in the back of the PDF version of the book)
22.5 Voltage for nonuniform fields by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
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