Transcript It is sometimes difficult to find the polarity of an induced emf. The net

Chapter 18.2
Capacitance
and Potential
1. A 5 μF capacitor is
connected to a 12 volt
battery. What is the
potential difference
across the plates of the
capacitor when it is
fully charged?
12 volts, a charged
capacitor has the
same potential
difference across the
plates as the source.
2. What is the charge on
one plate of the capacitor
in problem 1 when it is
fully charged? What is the
net charge on the
capacitor when it is fully
charged?
q = VC
q = 12 x 5 = 60 μC
or
q = 12(5 x
-6
10 )=
6x
-5
10
C
But the net charge on a capacitor is
always zero because the + and – plate
have charges of equal magnitude.
3. How much
electrical potential
energy is stored in
the capacitor in
problem 1?
2
CV
PEE = ½
2
PEE = ½ 5(12)
PEE = 360 μJ
or,
-6
2
PEE = ½ (5x 10 )(12)
-4
PEE = 3.6 x 10 J
4. If the plates are
separated by a
distance of 0.003 m,
what is the strength of
the electric field across
the plates?
Ed = V
E(0.003) = 12
E = 4000 N/C
5. A 9 μF capacitor is
connected to a 9 volt
battery. What is the
potential difference
across the plates of the
capacitor when it is
fully charged?
9 volts, a charged
capacitor has the
same potential
difference across the
plates as the source.
6. What is the charge on
one plate of the capacitor
in problem 5 when it is
fully charged? What is
the net charge on the
capacitor when it is fully
charged?
q = VC
q = 9 x 9 = 81 μC
or
q = 9(9 x
-6
10 )=
8.1 x
-5
10
C
But the net charge on a capacitor is
always zero because the + and – plate
have charges of equal magnitude.
7. How much
electrical potential
energy is stored in
the capacitor in
problem 5?
2
CV
PEE = ½
2
PEE = ½ 9(9)
PEE = 364.5 μJ
or,
-6
2
PEE = ½ (9 x 10 )(9)
-4
PEE = 3.645 x 10 J
8. If the plates are
separated by a
distance of 0.005 m,
what is the strength
of the electric field
across the plates?
Ed = V
E(0.005) = 9
E = 1800 N/C
9. List four ways
to increase the
charge on a
capacitor.
1. Larger plates
2. Plates closer
together
3. Different dielectric
4. Increase voltage.
10. What is the
difference in
electrical potential
energy and electric
potential?
Electric potential is
electric potential
energy divided by
charge.
11. A proton is moved
0.5 m in the direction of
an electric field with a
strength of 5000 N/C.
What is the change in
electric potential?
Ed = V
5000 x 0.5 = V
2500 V = V
Moving in the direction of the field
is a decrease of potential, so
-2500 V = V.
12. What is the
change in electrical
potential energy of
the proton in
problem 11?
PEE = qV
PEE = (1.6 x
-19
10 )(2500)
PEE = -4 x
-16
10
J
13. An electron is
moved 0.8 m in the
direction of an electric
field with a strength of
3000 N/C. What is the
change in electric
potential?
Ed = V
3000 x 0.8 = V
2400 V = V
Moving in the direction of the field
is a decrease of potential, so
-2400 V = V.
Remember! Potential (V) isn’t
affected by whether the charge
moving is positive or negative.
Potential only refers to what
would occur to a positive
charge, so we don’t even
consider the fact that the
electron is negative!
14. What is the
change in electrical
potential energy of
the electron in
problem 13?
V = EPE/q, so qV = EPE
(1.6 x10-19)2400 = 3.84 x 10-16 J
Since this is energy, it does
matter that it is an electron
(negative charge)! The electron
does not want to move in the
direction of the field, so the
answer is +3.84 x 10-16 J
15. A helium nucleus
(composed of two protons
and two neutrons) is moved
24 cm in the direction of an
electric field with a strength
of 9000 N/C. What is the
change in electric potential?
Ed = V
9000 x 0.24 = V
2160 V = V
Moving in the direction of the field
is a decrease of potential, so
-2160 V = V.
16. What is the
change in electric
potential energy of
the helium nucleus in
problem 15?
V = EPE/q, so qV = EPE
2(1.6 x10-19)2160 = 6.9 x 10-16 J
Since there are two protons, we
must multiply the proton charge
by 2.
The proton moves in the
direction of the field, so the
answer is -6.9 x 10-16 J