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
`DeltaPE=-∫Fdx`, [one dimension]
so for electric fields we divide by `q` to find
`DeltaV=-∫Edx`, [one dimension]
Applying the fundamental theorem of calculus yields
`E=-(dV)/(dx)` [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|=(kQ)/r^2`
The difference in voltage between two points on the same radius line is
`DeltaV=∫dV`
`=-∫E_xdx`
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.
`DeltaV=int_(r_1)^(r_2)E_rdx`
`=int_(r_1)^(r_2)(kQ)/r^2dr`
`=(kQ)/r]_(r_1)^(r_2)`
`DeltaV=(kQ)/r_2-(kQ)/r_1`.
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 `E_x =-(dV)/(dx)`.
(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.