# Episode 409: Uniform electric fields

So far we have mainly concentrated on the non-uniform fields around point or spherical charges. We will now discuss the physics of the uniform electric field, such as that between 2 parallel charged plates. The basic definitions of field strength and potential lead to different results than those for the non-uniform field.

**Summary**

- Discussion: Uniform electric fields (5 minutes)
- Demonstration: Potential and field strength in a uniform field (25 minutes)
- Discussion: Accelerating charges through a potential difference (10 minutes)
- Student questions: Uniform electric fields (10 minutes)
- Student questions: Millikan’s oil drop experiment (10 minutes)
- Discussion: Comparison of gravitational and electric fields (10 minutes)

**Discussion: Uniform electric fields**

What do we mean by a “uniform” field? (One that does not vary from place to place. In terms of the field lines, this means that they are parallel and evenly spaced. Also as field strength = – (potential gradient), the equipotentials should also be evenly spaced.)

Where have we seen such as field? (The field between two parallel charged plates.)

How will a charge move in such a field? The force is given by

*F* = *EQ*

Since *E* is constant, the force will be constant and therefore, b*y*

*F* = *ma*

the acceleration will also be constant and directed along the field lines for a positive charge and opposite to the field lines for a negative charge. (Note: A constant acceleration produces a parabolic path for the charge if it has any component of motion across the field lines, just as with projectiles in the Earth’s constant gravitational field near its surface).

**Demonstration: Potential and field strength in a uniform field**

A flame probe can be used to test the potential in the uniform field. Also, plotting a graph of potential against distance between the plates gives us the field strength because

*E* = – (potential gradient)

This demo will confirm the fact that the equipotentials are evenly spaced and parallel to the plates, giving a uniform electric field perpendicular to the plates.

Episode 409-1: Potential and field strength in a uniform electric field (Word, 65 KB)

Episode 409-2: Flame probe construction (Word, 237 KB)

Now, since *E* = – (potential gradient), but the potential gradient is uniform as we have seen, the electric field strength will simply be given (in magnitude) by:

where:

*V*= potential difference between plates (in volts)*d*= distance between plates (in metres)

The direction of the field is obviously from the positive to the negative plate.

**Discussion: Accelerating charges through a potential difference**

What happens when a charge is accelerated through a potential difference? If there is a potential difference between two points, there must be a field between them since *E* = - (potential gradient). Thus a charge that moves between the two points under the action of the field has work done on it by the field (as work = force ´ distance and a force is provided by the field). For a potential difference of *V* between two points, the work done by the field on a charge *Q* as it moves between those points is given by:

*W* = *VQ* where *W* is work done in joules, J.

(Careful here – so far we have used *V* as the symbol for potential, and V as the unit for potential, volts. Now we are using it for a potential *difference*. If this is likely to be a source of confusion, you may wish to adopt terminology like *W* = *Q* D*V* with D meaning “change in”).

We can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it.

(Note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 V,

i.e. W = 1 eV = 1.6 ´ 10^{-19} C ´ 1 V = 1.6 ´ 10^{-19} J).

In fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. In this case, *V* = *W*/*Q* is simply stating that the change in potential energy (or work done) per unit charge is equal to the change in potential (or potential difference), *V*.

**Student questions: Uniform electric fields**

Episode 409-3: Uniform electric fields (Word, 43 KB)

**Worked examples: Millikan’s oil drop experiment**

Episode 409-4: Millikan’s oil drop experiment (Word, 42 KB)

**Discussion: Comparison of gravitational and electric fields**

We finish this topic by discussing the parallels between gravitational and electric fields. You could ask your students to show the parallels in their own way – a useful summarising activity. Here are the main points that they should come up with:

We have seen the following similarities:

- Both fields follow an inverse square law for both force and therefore field strength
- For both types of field, potential and potential energy are inversely proportional to distance
- We define the zero of potential to be at infinity
- For attractions (and therefore always for gravitational fields) potential energies are always negative
- The major difference is that repulsions occur in electrostatic fields between charges of the same sign, whereas as far as we know, gravity is always attractive

The similarity of the fields may be brought home further by comparing the forms of the equations:

Gravitational fields | Electric fields |

g = –dV/dr | E = –dV/dr |

F = Gm_{1}m_{2}/r^{2} | F = kQ_{1}Q_{2}/r^{2} |

g = GM/r^{2} (a non-uniform field) | E = kQ/r^{2} (a non-uniform field) |

GPE = –Gm_{1}m_{2}/r | EPE = kQ_{1}Q_{2}/r |

V = –GM/r | V = kQ/r |

field strength = -(potential gradient)

**Download this episode**

Episode 409: Uniform electric fields (Word, 369 KB)