Transcript Physics_A2_31_CapacitorsEnergyStored

Book Reference : Pages 96-97
1.
To understand that when a capacitor is
charged it stores energy
2.
To be able to calculate the amount of energy
stored
3.
To be able to solve problems involving the
energy stored in a capactitor
We have seen from last lesson that when capacitor is
charged, the pd across the plates increases in proportion
to charge.
Consider one step in
the charging process a
capacitor from q to q +
q.
Plate pd / V
V
Work must be done to
force the extra charge
onto the plates & this
can be given by
v
q
Charge on plates / C
q + q
Q
E = vq
Where v is the average
voltage during this step
The work done in this small step from charging from q to
q + q is shown by the green vertical strip in the graph
If we consider all the
steps from a pd of 0 to
a the full pd of V then
the energy stored is
given by the area
under the graph.
Energy stored =
E= ½QV
Plate pd / V
This can be calculated
from ½ base x height
V
v
q
Charge on plates / C
q + q
Q
The energy equation :
E= ½QV
Can be re-written in alternative forms...
Since Q = CV
Substitute for Q : E= ½CV2
Or
Substitute for V : E= ½Q2 / C
Doubling the charge also doubles the pd (voltage)
& vice versa. If we look at the
E= ½CV2
Form of the equation we can see that this will
quadruple the energy stored
While charging the power supply (battery)
transfers energy equal to QV. Half of this energy
is stored in the capacitor (E= ½QV) the other half
is wasted through the resistance in the circuit and
dissipated to the surroundings as heat
Calculate the charge and the energy stored if a 10
F capacitor is charged to
3.0V and 6.0V
[30C & 45J, 60C & 180J]
In the following circuit a charged capacitor with
value C is allowed to discharge through a light
bulb.
Charge
Discharge
Readings :
C
V
In
Out
Switch
Joulemeter
Results :
Compare the difference
between the two joulemeter
readings with E= ½CV2
1. Charge the capacitor,
read the pd and the
initial joulemeter
reading
2. Discharge the capacitor
& take a further
joulemeter reading
-
-
-
-
+
+
+
+
During a storm the cloud
and the Earth below act
like a pair of charged
plates & a strong electric
field exists
If the cloud and ground are separated by d, then
then potential difference V is given by :
V = Ed
Where E is the electric field strength
For a cloud carrying a charge of Q then the
energy is given by ½QV which can be expanded
to ½QEd
If the wind moves the cloud to a new height d’
then the new energy stored will be ½QEd’
Note : The electric field strength E remains unchanged since it
depends on charge per unit area
The increase in energy is given by ½QEd’ - ½QEd
We can rewrite this as ½QEd where d = d’ – d
This increase in stored energy has come from the
work done by the wind overcoming the electrical
attraction between the cloud and Earth which
have opposite charges
The insulating properties of air break down
when the field strength reaches more than
about 300kV/m
A 50,000F capacitor is charged from a 9V
battery and then discharged through a light
bulb in a flash which lasts 0.2s. Calculate :
The charge and energy stored before discharge
The average power supplied to the light bulb
[0.45C & 2J, 10W]