Flashcards in 11.4 Capacitance Deck (15):

1

## What is capacitance?

### The charge per unit voltage that can be stored on a capacitor.

2

## What is the formula for capacitance?

###
C = Q/V

where Q is the charge on one of the plates and V is the potential difference between the plates.

3

## What is the unit for capacitance?

### Farad or Cv-1

4

## What is the formula for the energy stored in a capacitor?

### E=Q^2/2C = 1/2CV^2 = 1/2QV

5

## How is capacitance in series found?

### 1/Ct = 1/C1 + 1/C2 ....

6

## How is capacitance in parallel found?

### Ct = C1 + C2 ....

7

## What is the time constant?

### The time after which the stored charge has decreased to about 37% of its initial value.

8

## How is the capacitance for a parallel plate capacitor found?

###
C = E(A/d)

where A is the area of one of the plates

d is the separation of the plates

E is the permittivity of the medium between the plates.

9

## What happens if an insulator (dielectric) with permittivity of E is inserted between the plates of a capacitor?

###
The capacitance will increase because

E > E0

10

## Why does the capacitance increase?

### There is an electric field in between the plates directed from positive to negative. This electric field acts on the electrons of the dielectric, pulling them somewhat against the field, this is upwards. So there is separation of charge in the dielectric. This creates a small electric field within the dielectric this is directed upwards. This means that the net electric field in between the parallel plates is reduced compare to that in vacuum. Given by W = Fd = qEd and since E is reduced so is the work done, but the work done is also equal to W=qV this implies that the potential difference across the plates has been reduced. From the definition C = Q/V it follows that the capacitance increases.

11

## What is the charge on a capacitor and how can this lead to the equation for capacitance in a series circuit?

###
In a series circuit the potential difference is V, the charge of the first capacitor is q1 and on the second in q2. We have that q1 = C1V and q2 = C2. The total charge on the two capacitors is q = q1 + q2 = (C1 + C2)V. We man define the total capacitance of the combination used q = C total V so

C total = C1 + C2.

12

## What is the charge on a capacitor and how can this lead to the equation for capacitance in a parallel circuit?

###
The charge on each capacitor is the same, (if it were not the segment between the two capacitors would not have net charge), we know that V1 = q/C1 and V2 = q/C2 and that the source of the potential difference V is equal to V1 + V2 the total capacitance is then V = q/Ctotal = q/C1 + q/C2 giving

1/Ctotal = 1/C1 + 1/C2

13

## What happens to the capacitor while charging?

### Initially the capacitor is uncharted. As soon as the switch is moved to a position where it can charge, the charge on the capacitor plates will increase to a final contact value, the potential difference across the capacitor plates will increase to the EMF of the battery, the current starts out large and then decreases exponentially. The capacitor is now fully charged and the current become zero. The final charge on the plates q0=CE. The initial current is equal to I0 = E/R.

14

## What happens to the capacitor while discharging?

###
Initially the capacitor is fully charged. As soon as the switch is moved to position where it can discharge the current, charge and voltage all decrease exponentially according to

q =qoe^(-t/RC) V=Voe^(-t/RC) I=(Vo/R)e^(-t/RC)

Qunaitites with a subscript of 0 indicate values t t=0. The quantity RC is called the time constant and is donated by T. After a time t=T the charge will be

q=qoe(-t/T)=qo/e = 37% of its initial value.

15