Capacitors in circuits
Capacitors, capacitance.
Circuits with capacitors
A parallel plate capacitor consists of two
metal plates, one carrying charge +q and
the other carrying charge –q.
It is common to fill the region between
the plates with an electrically insulating
substance called a dielectric.
Quick lab: charge the capacitor with a hand generator and measure
potential difference across it.
Make a circuit with: capacitor, light bulb, switch and A-meter. Keep the
switch open. Get the timer ready to measure the time it will take for the
current to die out. Complete the circuit, measure the time and sketch a
graph I(t) in the space below (use your actual numbers to mark the
significant points).
THE RELATION BETWEEN CHARGE AND POTENTIAL
DIFFERENCE FOR A CAPACITOR
The magnitude of the charge in each place of the
capacitor is directly proportional to the magnitude
of the potential difference between the plates.
q  CV
The capacitance C is the proportionality constant.
SI Unit of Capacitance: coulomb/volt = farad (F)
Charge the capacitor again, use V-meter to determine the charge on
either of the plates of the capacitor
THE DIELECTRIC CONSTANT
If a dielectric is inserted between the plates of
a capacitor, the capacitance can increase markedly.
Dielectric constant
Eo

E
THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR
q
Co =
V
V = Eod
Eo = q ( e o A )
Eo
E=
k
q
C=
Eo d
qe o A e o A
C=
=
qd
d
C=
ke o A
d
Three ways to increase capacitance:
A 5.0 F capacitor is connected to a 6.0V battery and then to a
12.0V battery. If C = Q/V, will the capacitance increase or
decrease?
Example 12 A Computer Keyboard
One common kind of computer keyboard is based on the
idea of capacitance. Each key is mounted on one end
of a plunger, the other end being attached to a movable
metal plate. The movable plate and the fixed plate
form a capacitor. When the key is pressed, the
capacitance increases. The change in capacitance is
detected, thereby recognizing the key which has
been pressed.
The separation between the plates is 5.00 mm, but is
reduced to 0.150 mm when a key is pressed. The
plate area is 9.50x10-5m2 and the capacitor is filled with
a material whose dielectric constant is 3.50.
Determine the change in capacitance detected by the
computer.
C  19.0 10 12 F
ENERGY STORAGE IN A CAPACITOR
EPE = 12 CV 2
Produce two other formulas, excluding:
a) C
b) V
q  q1  q2  C1V  C2V  C1  C2 V
Parallel capacitors
CP  C1  C2  C3  
1
q
q
1 

V  V1  V2  
 q  
C1 C2
 C1 C2 
Series capacitors
1
1
1
1




CS C1 C2 C3