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22.5 Voltage for nonuniform fields by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.

vCalc Companion Formulas | |

vCalc Formulary | 22.5 Voltage for nonuniform fields |

`E=-(dV)/(dx)` | Electric Potential |

`V=(kQ)/r` | Voltage at distance `r` |

`DeltaV=(kQ)/r_2-(kQ)/r_1` | Change in voltage |

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=??_(r_1)^(r_2)E_rdx`

`=??_(r_1)^(r_2)(kQ)/r^2dr`

`=(kQ)/r]_(r_1)^(r_2)`

The standard convention is to use `r_1=?` 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.