PhysicsLAB: Parallel Plate Capacitors

When two parallel plates are connected across a battery, the plates will become charged and an electric field will be established between them.

Remember 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 is calculated with the equation E = V/d.

Note that the electric field strength, E, can be measured in either the units V/m, or equivalently, N/C.

[E] = V/d
(J/C)/m
(Nm)/C/ m

N/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. That force is calculated with the equation F = qE. To review more about electric fields between parallel plates, go back and review this resource lesson.

Capacitance

When two plates are charged and used in an electric circuit, that device is called a capacitor. It's role in the circuit is to store energy. Capacitors are rated in terms of their capacitance which is measured in farads (F). One farad equals the ratio of one coulomb per volt.

[F] = C/V

A parallel plate capacitor's effective capacitance is defined in terms of its geometry.

C = ε_oA/d

where

ε_o, the permittivity of free space, is a constant equal to 8.85 x 10^-12 F/m,
A is the cross sectional area of ONE plate, and
d is the distance between the plates.

Essentially, capacitance measures the relative amount of charge that can be stored on a pair of parallel plate for a given amount of voltage. If the capacitance increases, then more charge can be stored when the same potential is applied.

The equation for the line becomes Q = CV and the equation for the area under the curve becomes E = ½QV = ½CV².

The plates can then be discharged later through an external circuit. They are used when the circuit requires a big burst of energy; for example: to "jump start" electric motors, TV's or operate flash attachments on a camera.

Refer to the following information for the next three questions.

A parallel-plate capacitor is connected across a 9-volt battery. Each plate of the capacitor has a cross-sectional area of 0.0016 m² and the plates are separated by 5 µm of air.

	What is the capacitance of this capacitor?

	How much charge is present on each plate?

	How much energy is stored in the capacitor?

Combinations of Capacitors

When more than one capacitors are used in a circuit, the above formula is restated as

Q_total = C_total x V_total

If the capacitors are arranged in series (one after another along a single path), then

Q_series = Q₁ = Q₂ = Q₃
C_series = (1/C₁ + 1/C₂ + 1/C₃)^-1
V_series = V₁ + V₂ + V₃

If the capacitors arranged in parallel (strung along multiple paths that cross the same section), then

Q_parallel = Q₁ + Q₂ + Q₃
C_parallel = C₁ + C₂ + C₃
V_parallel = V₁ = V₂ = V₃

Springs and Capacitors

Let's take a moment and note a similarity between springs and capacitors.

For a simple spring, F_distorting = ks and the energy stored is PE_e = ½ks².

When springs are combined in series, the spring constant for the system becomes

k_series = (1/k₁ + 1/k₂ + 1/k₃)^-1.

When springs are combined in parallel, the spring constant for the system becomes

k_parallel = k₁ + k₂ + k₃

These rules exactly model those of capacitors. The similarities make sense since both springs and capacitors are energy-storage devices: springs store mechanical energy; capacitors store electrical energy.

Resistors and Capacitors

Note that there are both similarities and differences between the rules for capacitors and resistors.

When resistors are wired in series,

I_series = I₁ = I₂ = I₃

R_series = R₁ + R₂ + R₃

V_series = V₁ + V₂ + V₃

Notice that these "circuit properties" agree with those of capacitors:

the current (C/sec) through devices wired in series is the same, charges will "flow" between capacitors until they equalize;
voltages changes across devices wired in series are additive, whether they are resistors or capacitors.

Similarly, for resistors wired in parallel,

I_parallel = I₁ + I₂ + I₃

R_parallel = (1/R₁ + 1/R₂ + 1/R₃)^-1

V_parallel = V₁ = V₂ = V₃

Notice once again the agreement of these "circuit properties" with those of capacitors:

when currents (C/sec) divide in parallel, the charges on the capacitor plates would consequently need to add together;
the voltage across parallel devices is the same, whether they are resistors or capacitors.

However, the rules for resistance and capacitance are "reversed" since resistors are devices that dissipate energy while capacitors are devices that store energy.

Practice Examples

Refer to the following information for the next six questions.

Each of the following capacitors has a rated capacitance of 10 µF.

	Are these capacitors arranged in parallel or series?

	How much voltage is across each capacitor?

	How much charge is stored on each capacitor?

	Which side of the capacitor is positive? Negative?

	How much total charge is stored on the entire set of capacitors?

	What is the total capacitance of this collection?

Refer to the following information for the next five questions.

Each of the following capacitors has a rated capacitance of 10 µF.

	Are these capacitors arranged in parallel or series?

	What is the total capacitance of this collection?

	How much total charge is stored on the entire set of capacitors?

	Which side of the capacitor is positive? Negative?

	What voltage is across each capacitor?

Adjustable Parallel-Plate Capacitors

In the following examples we will use an adjustable, parallel-plate capacitor that can be switched between two positions, A and B, without disturbing the electrical circuit:

in position A, the geometry of the plates creates a capacitance of 4 x 10^-9 F,
in position B the geometry dictates a new capacitance of 3.7 x 10^-9 F.

Refer to the following information for the next four questions.

While in position A, the capacitor is initially charged by a 12-V battery. The battery is removed and the capacitor is moved from position A to position B without changing the charge on its plates. Since the battery is removed during the transition from position A to position B, the voltage will be permitted to change even though the charge on the plates must remain constant.

	How much charge is on the capacitor in position A?

	What is the voltage across it in position B?

	How much electrical energy is stored in each position?

	What is the minimum amount of work that a person holding the plates must have done to adjust the capacitor from position A to position B?

Refer to the following information for the next four questions.

While in position A, the capacitor is initially charged by a 12-V battery. The battery remains attached to the plates as the capacitor is moved from position A to position B. Since the battery is NOT removed during the transition from position A to position B, the voltage across the plates must remain constant since it is regulated by the battery's presence in the circuit. This time the charge on the plates will be permitted to change.

	What is the voltage across it in position B?

	How much charge is on the capacitor in positions A and B?

	How much electrical energy is stored in each position?

	What is the minimum amount of work that a person holding the plates must have done to change the capacitor from A to B?