Download Two conductors in proximity form a “capacitor”

Two conductors in proximity form a “capacitor”: they have a
capacity to hold a charge Q (+Q on one and -Q on the other)
with a voltage difference V.
C=Q/V
Note: Capacitance C is italicized. Coulomb C is not. (Sorry!)
Capacitance is a property of the geometry of the conductors
(and depends on whether there is vacuum or material
between them, as we will see later.)
This is because of linearity (superposition.) If I double the
source charges, I will double the field, and double the
potential.
If I put ±2 microcoulombs of charge on the sphere and plate,
there is a potential difference of 2 V.
What is the capacitance?
A] 0
B] Cannot determine, since not symmetric
C] 1 microfarad = 10-6 C/V
D] 106 farads
If I put ±2 microcoulombs of charge on the sphere and plate,
there is a potential difference of 2 V.
If I put ±4 microcoulombs of charge on the sphere and plate, what
will be the potential difference?
A] 0
B] 2 V
C] 4 V
D] 8 V
Two infinite parallel sheets carry charge densities ± 
What is the electric field at point 1?
Two infinite parallel sheets carry charge densities ± 
What is the electric field at point 2?
The potential difference between the plates is d
By superposition, the field in the dielectric is the sum of the
original capacitor field, plus the field from the polarization
surface charge.
The total field is therefore
A] bigger than the original capacitor field
B] smaller than the original capacitor field
C] the same as the original capacitor field
Answer: smaller than the original field.
We can write that the new field is the original field, divided by
K, the “dielectric constant”
• What is the new potential across the capacitor?
A] same as the old potential, without the dielectric
B] bigger than the old potential, by a factor of K
C] smaller than the old potential, by a factor of K
If I put a charge ±Q on a parallel plate capacitor and
increase the plate separation, what happens to V?
A] it increases
B] it decreases
C] it stays the same
If I put a charge ±Q on a parallel plate capacitor and
increase the plate separation, what happens to V?
The field is the same, but the distance is larger. So V increases.
What happens to the potential energy U?
A] it increases
B] it decreases
C] it stays the same
If I put a charge ±Q on a parallel plate capacitor and
increase the plate separation, what happens to V?
The field is the same, but the distance is larger. So V
increases.
What happens to the potential energy U?
A] it increases
The energy density u is the same, but there is more volume.
Where did this energy come from??
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
A] it increases
B] it decreases
C] it stays the same
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
The field in the dielectric is reduced by a factor of K. So the potential goes
down.
What happens to the potential energy?
A] it goes up by a factor of K
B] it goes up by a factor of K2
C] it goes down by a factor of K
D] it goes down by a factor of K2
E] it stays the same.
If I put a charge ±Q on a parallel plate capacitor and insert a
dielectric, what happens to V?
What happens to the potential energy?
It may be counterintuitive, but the potential energy goes down by a
factor of K (not K2).
U=Q2/(2C); the capacitance goes up by a factor of K.
2
1
u


E
2
Note: the energy density for a given E field in a dielectric is
The field goes down by a factor of K, but epsilon adds a factor of K.

If the potential energy goes down, where does the energy go?
• The field pulls the dielectric into the capacitor, giving it
kinetic energy. (Very small, here.)
• Application: Optical tweezers
Batteries don’t store charge. They store energy.
A chemical battery works because electrons like to
leave some materials to go to others.
The change in energy when the electron goes “downhill”
becomes available as electric POTENTIAL between
the battery terminals
Lead Acid (Car) Battery
The rxn on the right will not go in
the direction indicated unless the
electrolyte sol’n potential is closer
than 1.685 V to the + electrode
potential.
i.e. the + electrode can be no more
than 1.685 V higher in potential
than the electrolyte sol’n.
Pb+2 is in the form of solid lead sulfate on
the electrodes. When “discharged”, both
electrodes turn into lead sulfate.
Lead Acid (Car) Battery
The rxn on the left will not go in the
direction indicated unless the
electrolyte potential is closer than
0.356 V to the - electrode potential.
i.e. the - electrode can be no more
than 0.356 V lower in potential than
the electrolyte soln.
Lead Acid (Car) Battery
Overall, then, the reactions will
STOP when the potential between
the +/- electrodes is 2.041V.
Only a tiny tiny tiny amount of
charge needs to build up on the
electrodes for the reaction to stop.
How much?
But if you connect the electrodes
with a resistive wire, the reaction
will start to go as the potential
drops a hair below 2.041V.
Lead Acid (Car) Battery
What happens if you force the
potential difference to be higher
than 2.041 V?
The reactions run backwards!
Lead sulfate turns into lead oxide
and lead (metallic).
This is charging the battery.
Lead Acid (Car) Battery
Note that the rxn doesn’t make a
big “reservoir” of electrons.
The battery doesn’t die because it
runs out of stored electrons.
It doesn’t store electrons.
It “pumps” electrons “on demand”,
i.e. when the potential falls below
2.041 V.
The two electrodes of an ideal V volt battery are shown. The field lines
for the electric field between the electrodes are shown when nothing is
attached to the electrodes. The electrodes have a separation = d.
A resistive wire of length L is then attached to the battery.
What is the electric field in the wire, when steady state is reached?
A] 0
B] it varies, but averages to V/d
C] it varies, but averages to V/L
D] it is V/L everywhere in the wire
E] no way to determine
The two electrodes of an ideal V volt battery are shown. The field lines for
the electric field between the electrodes are shown when nothing is
attached to the electrodes. The electrodes have a separation = d.
A resistive wire of length L is then instantaneously attached to the battery.
What is the electric field in the wire, immediately after attaching it?
A] 0
B] it varies, but averages to V/d
C] it varies, but averages to V/L
D] it is V/L everywhere in the wire
E] no way to determine
Where on this wire will negative charges “build up” (a tiny amount)
B
A
Assume wires have zero resistance
Assume wires have zero resistance
No E field in electrostatics.
A] same
B] 3R
C] 9R
D] R/3
E] R/9
On left, both are equally bright. Bulb A on the right is brighter,
As V^2/R is bigger.
If the R’s are all the same,
which arrangement has the
lowest Ref ? B
Which has the highest Ref ?
A
Which has a lower Ref:
C or d? D
If the 2 cm long resistor is 3 ohms, what is the resistance of the 1 cm
long resistor (with half the radius)? (Both are of the same material.)
A] 1 ohm
B] 1.5 ohms
C] 2 ohms
D] 3 ohms
E] 6 ohms
What is the current through the battery?
A] 0 amps
B] 0.5 amps
C] 1 amp
D] 2 amps
E] 3 amps
What is the power delivered by the battery? In watts
A] 0
B] 1
C] 2
D] 3
E] 4
What is the magnitude of the E field the physically larger resistor (3
ohms)? (in V/m)
A] 0
B] 100
C] 200
D] 300
E] 400
What is the magnitude of the E field the physically smaller resistor (6
ohms)? (in V/m)
A] 0
B] 100
C] 200
D] 300
E] 400
What is the current through the
battery? In amps
A] 0
B] 1/9
C] 1/6
D] 2/9
E] 2/3
What is the E field in the 3 ohm
resistor? (in V/m)
A] 0
B] 16
C] 33 1/3
D] 78
E] 200
Are R1 and R2 in parallel, in series, or neither?
A] parallel
B] series
C] neither
Which junction formula is correct?
Which loop formula is correct?
At steady state, what is current I1?
At steady state, if the voltage at a = 0, what is the
voltage at b?