CK-12-Physics - Intermediate

http://www.ck12.org Chapter 16. Electric Potential

Experiments show that if the chargeQand voltageVof a capacitor are measured, along with its areaAand separation
d, a constant of proportionality (denoted as the Greek letter Epsilonε 0 ) is found by considering that

C∝

A

d

→

Q

V

∝

A

d

→

Q

V

=ε 0

A

d

→ε 0 =

Qd

VA

Epsilon(ε 0 )is called the permittivity of free space and has units

C

V∗m→

C^2
N∗m^2.
The latter set of units can be shown to be the reciprocal of Coulomb’s constantk, and the numerical value of Epsilon
is

ε 0 =

1

4 πk

= 8. 85 × 10 −^12

C

N∗m^2

Notice that the capacitanceCis independent of the charge or voltage of the capacitor. It is only dependent upon
the physical dimensions of the capacitor, and, as we shall see, upon the type of the insulator placed between its
conducting plates.

Check Your Understanding

1. The plates of a capacitor are 1.5 mm apart. What area must the plates have in order for the capacitance to be 1.0
   F?

Answer:

C=ε 0

A

d

→A=

Cd

ε 0

→A=

( 1. 0 F)( 1. 5 × 10 −^3 mm)

8. 85 × 10 −^12 C

2

N∗m^2

= 1. 69 × 108 → 1. 7 × 108 m^2

Assuming the plate is a square, it would have a side length of 13 km.

One farad is, indeed, a very large capacitance. We often find it more practical to express capacitance in terms of
microfarads(μF→ 10 −^6 F)and picofarads(pF→ 10 −^12 F).

2. The plates of a capacitor have dimensions 3.00 cm by 2.00 cm and are separated by a distance of 1. 00 × 10 −^3 m.
   What is the capacitance of the capacitor?

Answer:

C=ε 0

A

d

= ( 8. 85 × 10 −^12 )

( 3. 00 × 10 −^2 m)( 2. 00 × 10 −^2 m)

1. 00 × 10 −^3 m

= 5. 31 × 10 −^12 → 5. 31 pF

3. The capacitance of a parallel plate capacitor is 250μF.

a. What is the maximum amount of charge deposited on each plate if the capacitor is connected to a 9.0 V battery?

Answer:

Q=CV→Q= ( 250 × 10 −^6 F)( 9. 00 V) = 2. 25 × 10 −^3 C