 Electric Fields: Parallel Plates Printer Friendly Version
As shown below, when two parallel plates are connected across a battery, the plates become charged and an electric field is established between them. In this diagram, the battery is represented by the symbol where the longer line is the battery's positive terminal and the shorter line is its negative terminal. Batteries are rated by to their volts where 1 volt is defined as 1 joule per coulomb or 1 V = 1 J/C. This tells us that a 1.5 V battery can "energize" 1 µC of charge by 1.5  µJ. The difference between a 1.5-V AAA battery and a 1.5-V D-cell battery lies in their power. A D-cell can supply its energy for a longer amount of time.

Recall that the direction of an electric field is defined as the direction that a positive test charge would move. So in this case, the electric field would point from the positive plate to the negative plate. Since the field lines are parallel to each other, this type of electric field is uniform and has a magnitude which can be calculated with the equation E = V/d where V represents the voltage supplied by the battery and d is the distance between the plates.

It should now be noted that there are two units in which the electric field strength, E, can be measured: either in V/m, or equivalently, in N/C. That is, expressing the electric field strength as E = 10 V/m is equivalent to stating that the electric field strength equals 10 N/C.

V/d = (J/C)/m
[(nt m)/C]/m
nt/C

Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field. That force is calculated with the equation F = qE where both F and E are vector quantities and q is a scalar quantity. In the diagram above, the distance between the plates is 0.14 meters and the voltage across the plates is 28V. The magnitude of the UNIFORM electric field between the plates would be

E = − ΔV/d
= − (-28 V) / (0.14 m)
= 200 V/m

If a positive 2 nC charge were to be inserted anywhere between the plates, it would experience a force having a magnitude of

F = qE
= (2 x 10–9 C) (200 N/C)
= 4.0 x 10-7 newtons

towards the negative, or bottom plate, no MATTER where it is placed in the region between the plates.

A volt is a scalar quantity that equals a joule per coulomb. Based on this definition, moving a coulomb of charge across a potential difference of 1 volt would require 1 joule of work. Surfaces that have the same potential, or voltage, are called equipotential surfaces and are presented by dotted lines that are always drawn at right angles to the field lines (the solid vectors). Field lines always point from regions of high potential to regions of low potential. In our diagram, the top plate would be at +28 V and is the "high potential plate" while the bottom plate would be at 0 V and is the "low potential plate."

When analyzing electric fields between parallel plates, the equipotential surfaces between the plates would be equally spaced and parallel to the plates. In the diagram shown, we have drawn in six equipotential surfaces, creating seven subregions between the plates. The distance from one surface to another would equal 0.14/7 or 0.02 meters. Therefore the potential difference from one equipotential surface to the next would equal

ΔV = − Ed cos θ

In our example θ = 0º since our 2 nC positive charge will be moving in the same direction as the field lines; that is, towards the negative plate.

ΔV = − (200 V/m)(0.02 m)
ΔV = − 4 volts

Numbering successively from the top plate (+28 V) to the bottom plate (0 V), our equipotential surfaces would have voltages of 24 V, 20 V, 16 V, 12 V, 8 V, and 4 V respectively. In general, notice that

• when traveling in the direction of the electric field, ΔV would be negative since θ equals 0º and cos 0º = 1. Also note that traveling along a field line always moves the charge from positions of high potential to those of lower potential.
• when traveling against the field lines, ΔV would be positive since θ equals 180º and cos 180º = -1. Also the charge would be moving towards a position of higher potential.

Charges possess electric potential energy (EPE) based on their position in an electric field just like masses possess gravitational potential energy (PE) based on their position in a gravitational field.

PE = mgh
EPE = qV

Summarizing what we have learned so far. Our 2 nC charge, no matter where it is placed in the electric field, will always experience a force of 4.0 x 10-7 newtons towards the bottom, negative plate. That constant force would result in the charge being uniformly accelerated towards the bottom plate. As the charge travels downward along any arbitrary field line, it will move between equipotential surfaces and lower its electric potential energy (EPE) from

2 x 10-9 x (28) = 56 x 10-9 J to
2 x 10-9 x (24) = 48 x 10-9 J to
2 x 10-9 x (20) = 40 x 10-9 J to
2 x 10-9 x (16) = 32 x 10-9 J to …… eventually 0 J

Recall that work done ( Wby the field = Fs cos θ where θ = 0º) by a conservative field lowers a particle's potential energy. In this case, it is lowering the charge's electric potential energy (EPE). But remember that the field is also accelerating the charge and thus increasing its kinetic energy (KE).

Wby the field = − ΔEPE = + ΔKE

The amount of work done on the 2 nC charge as it moves between each set of successive equipotential surfaces equals

Wby the field = Fs cos 0º
= (qE)s
= (2 x 10-9 C)(200 N/C)(0.02 m)
= 8 x 10-9 J

Wby the field = Fs cos 0º
= − ΔEPE
= − q(Vfinal - Vinitial)
= − (2 x 10–9 C)(-4 J/C)
= + 8.0 x 10-9 J

Applying conservation of energy, the electric potential energy lost by the charge will be equal to the kinetic energy it gains. This means that the 2 nC charge would gain 8.0 x 10-9 J of KE as it moves between each successive set of equipotential surfaces towards the negative plate. This is exactly what we would be expecting since the charge is being accelerated towards the negative plate.

Note that NO work would be done if the 2 nC charge were to be moved along an equipotential surface since all points on a surface are at the same potential.

Refer to the following information for the next question.

In the following diagram, the plates are connected across a 60 V power supply and are separated by 2 cm. How fast would an electron, if released from rest next to the negative plate, hit the upper positive plate?

Refer to the following information for the next question.

In the following diagram, the plates are connected across a 60 V power supply and are separated by 2 cm. Note that the electron's initial trajectory places it midway between the two plates. How fast and at what angle would an electron initially moving horizontally at 3 x 106 m/sec strike the upper positive plate?

Refer to the following information for the next two questions.

A charged ball, of mass 10 grams and charge -6 µC, is suspended between two metal plates which are connected to a 60 V power supply and are 2 cm apart.
 What is the angle at which the ball hangs?

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