Download Introduction to Capacitance Part A What property of objects is best

Introduction to Capacitance
Part A
What property of objects is best measured by their capacitance?
ability to store charge
Correct
Capacitance is a measure of the ability of a system of two conductors to store electric
charge and energy. Capacitance is defined as
. This ratio remains constant as long
as the system retains its geometry and the amount of dielectric does not change.
Capacitors are special devices designed to combine a large capacitance with a small size.
However, any pair of conductors separated by a dielectric (or vacuum) has some
capacitance. Even an isolated electrode has a small capacitance. That is, if a charge
placed on it, its potential
capacitance .
Part B
Assume that charge
with respect to ground will change, and the ratio
is placed on the top plate, and
is
is its
is placed on the bottom plate.
What is the magnitude of the electric field between the plates?
Express
in terms of and other quantities given in the introduction, in addition to
and any other constants needed.
=(q/(epsilon_0*A))Correct
Part C
What is the voltage
between the plates of the capacitor?
Express in terms of the quantities given in the introduction and any required physical
constants.
=((q*d)/(epsilon_0*A))Correct
Part D
Now find the capacitance
Express
of the parallel-plate capacitor.
in terms of quantities given in the introduction and constants like
=epsilon_0*(A/d)Correct
.
You have derived the general formula for the capacitance of a parallel-plate capacitor
with plate area and plate separation . It is worth remembering.
Part E
Consider an air-filled charged capacitor. How can its capacitance be increased?
Decrease the spacing between the plates of the capacitor.
Correct
Part F
Consider a charged parallel-plate capacitor. Which combination of changes would
quadruple its capacitance?
Halve the plate separation; double the plate area.
Correct
Finding the Capacitance
Part A
What is the capacitance
of the second capacitor?
Express your answer in terms of
,
, and constants.
=(K*C_1)/(K-1)Correct
As a check that this result makes sense, note that if
capacitor being replaced by a conductor, then
is still greater than 1),
, which corresponds to the first
. In the limit as
approaches 1 (but
becomes very large, effectively making the second capacitor a
conductor. This is also expected, since if
capacitor.
Force between Capacitor Plates
, then no change was made to the first
Part A
Find
, the magnitude of the force each plate experiences due to the other plate as a
function of , the potential drop across the capacitor.
Express your answer in terms of given quantities and .
=(A*epsilon_0*V^2)/(2*d^2)Correct
Capacitors with Partial Dielectrics
Part A
The horizontal capacitor is filled halfway with a material that has dielectric constant
.
What fraction of the area of the vertical capacitor should be filled (as shown in the
figure) with the same dielectric so that the two capacitors have equal capacitance?
Express your answer in terms of
.
=1/(1+K)Correct
Energy of a Capacitor
Part A
Find
, the the energy dissipated in the resistor.
Express your answer in terms of
and other given quantities.
=U/KCorrect
Can you explain where the rest of the energy went?
Part B
Consider the same situation as in the previous part, except that the charging battery
remains connected while the dielectric is inserted.
The battery is then disconnected
and the capacitor discharged. For this situation, what is
resistor?
, the energy dissipated in the
Express your answer in terms of
and other given quantities.
=U*KCorrect
Can you explain where the extra energy came from? (Answer below)
In this case, the extra energy comes from the battery which provides some extra charge,
to maintain the potential difference across the plates.
The first case is a little more involved. The dielectric experiences an attractive force as it
is inserted into the capacitor. Therefore it would either gain kinetic energy, or work
would have to be done against any resistive forces on it that prevent it from gaining
speed. This is where the the energy goes. You could actually calculate the force on the
dielectric as
, where
is the energy in the capacitor as a function of the
length of the dielectric inserted into the capacitor.