Transcript V 2

Voltage and Capacitance
Chapter 29
Electric Potential Energy
Potential Energy of a charge
• Wants to move when it has
high PE
• Point b
– U = max
– K = min
• Point a
– U = min
– K = max
Electric Work
Charge moving between plates.
Work = FDr cos0o = qExf - qExi
DU = qExf - qExi
A 2.0 cm X 2.0 cm capacitor has a 2.0 mm spacing
and a charge of +1.0 nC.
a. Calculate the electric field of the capacitor. (2.82 X
105 N/C)
b. A proton is released from rest at the positive plate.
Calculate the change in potential energy. (9.02 X
10-17 J)
c. Calculate the final speed of the proton. (3.29 X 105
m/s)
d. An electron is released from the halfway point
between the plates. Calculate the change in PE and
the final speed of the electron. (9.95 X 106 m/s)
A 2.0 cm diameter disk capacitor has a 2.5 mm
spacing and a charge of +0.75 nC.
a. Calculate the electric field of the capacitor.
b. An electron is released from rest at the negative
plate. Calculate the change in potential energy.
(9.02 X 10-17 J)
c. Calculate the final speed of the electron. (3.29
X 105 m/s)
Potential Energy of point charges
Uelec = 1 q1q2
4peo r
k = 9.0 X 109 Nm2/C2
A proton is fired from “far away” at a 1.0 cm
diameter glass sphere of charge +100 nC.
a. Calculate the initial potential energy of the
system (just as it touches the sphere).
b. Since PE = KE, calculate the needed initial
speed of the proton.
An electron and a positron are created in the CERN
collider.
a. Calculate the potential energy they have when
they are 1.0 X 10-10 m apart.
b. Calculate the velocity they need to escape from
one another. Remember that PE = KE, but you
will need to consider the KE of both particles
added together.
Electric Potential: Voltage
• Voltage
• 1 Volt = 1 Joule/Coulomb
V = DU
q
Vab = Va – Vb = -Wba
q
Work done by the
electric field to
accelerate the
charge
• The higher the rock, the greater the PE
• The greater the Voltage or charge, the greater the
PE (DU = qV)
An electron is accelerated in a TV tube through a
potential difference of 5000 V.
a. Calculate the change in PE of the electron (-8.0
X 10-16 J)
b. Calculate the final speed of the electron (m = 9.1
X 10-31 kg) (4.2 X 107 m/s (1/7th speed of light)
Calculate the final speed of a proton (mass =
1.67 X 10-27 kg) (9.8 X 105 m/s (0.3% speed of
light)
Equipotential Lines
• Equipotential lines are
perpendicular to electric field
lines
• Voltage is the same along
equipotential lines
• Like contour (elevation) lines on
a map
Equipotential lines for point charges
Electric Field and Voltage
U = qEs
V = DU/q
V = Es
Greater the distance between plates, the greater the
voltage
The greater the E field, the greater the voltage
Voltage increases as you move between plates
What is the electric field between two plates
separated by 5.0 cm with a voltage of 50V. (1000
V/m)
A capacitor is constructed of 2.0 cm diameter
disks separated by a 2.0 mm gap, and charged
to 500 V.
a. Calculate the electric field strength (V = Ed)
b. Calculate the charge on each plate (E = Q/e0A)
c. A proton is shot through a hole in the negative
plate towards the positive plate. It has an initial
speed of 2.0 X 105 m/s. Does is have enough
energy to reach the other side? (V = DPE/q)
Electron Volt
• Energy an electron gains moving through a
potential difference of 1 V
1 eV = 1.6 X 10-19 J
• Ex: An electron moving through 1000 V would
gain 1000 eV of energy
Voltage due to a Point Charge
• Voltage is not directional (scalar)
• Charged particles (i.e.: electrons, protons) have a
voltage
V = kQ
r
Example 1
Consider a +1.0 nC charge.
a. Calculate the electric potential (voltage) at a
point 1.0 cm from the charge
b. Calculate the electric potential at a point 3.0 cm
from the charge.
Point Charges: Example 2
Calculate voltage (electric potential) at point A as
shown below:
A
30 cm
52 cm
Q1 = +50 mC
Q2 = -50 mC
Use Pythagoream theorem to calculate the distance
from A to Q2:
A
30 cm
52 cm
Q1 = +50 mC
Q2 = -50 mC
VA = V1 + V2
V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)
r
(0.30 m)
V1 = 1.50 X 106 V
V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)
r
(0.60 m)
V2 = -7.5 X 105 V
VA = 1.50 X 106 V -7.5 X 105 V
VA = 7.5 X 105 V
Point Charges: Example 3
Calculate voltage (electric potential) at point B as
shown below:
B
30 cm
26 cm
Q1 = +50 mC
26 cm
Q2 = -50 mC
Use Pythagoream theorem to calculate the distance
from B to Q1 and to Q2:
B
30 cm
26 cm
Q1 = +50 mC
26 cm
Q2 = -50 mC
VA = V1 + V2
V1 = kQ = (9.00X 109 Nm2/C2)(5.00X10-5C)
r
(0.40 m)
V1 = 1.125 X 106 V
V1 = kQ = (9.00X 109 Nm2/C2)(-5.00X10-5C)
r
(0.40 m)
V2 = -1.125 X 105 V
VA = 1.125 X 106 V –1.125 X 105 V
VA = 0 V
Point Charges: Example 4
How much work is required to bring a charge of q =
3.00 mC to a point 0.500 m from a charge Q =
20.0 mC?
VQ = kQ
r
=
(9.00X 109 Nm2/C2)(2.00X10-5C)
(0.500 m)
VQ = 3.6 X 105 V (This is the voltage caused by
the stationary charge)
W = DU (like the work to lift a book to a shelf)
V = DU
q
V=W
q
W = Vq
W = (3.6 X 105 V )(3 X 10-6 C)
W = 1.08 J
Point Charges: Example 5
Which of three sets of charges has the most:
• positive potential energy?
• The most negative potential energy?
• Would require the most work to separate?
+
+
+
(i)
-
-
+
(ii)
(iii)
Charged Spheres
Act as point charges
At surface of sphere
Substitute:
A proton is released from rest at the surface of a
1.00 cm-diameter sphere charged to +1000 V.
a. Calculate the charge on the sphere (0.56 nC)
b. Calculate the speed of the proton when it is 1.00
cm from the sphere (note that it has potential
energy in both cases, U = qV) (3.6 X105 m/s)
A thin ring has a radius R and charge Q. Find the
potential at a distance of z from the axis of the
ring.
A thin disk has a radius R and charge Q. Find the
potential at a distance of z from the axis of the
ring.
A 17.5 mm diameter dime is charged to +5.00 nC.
a. Calculate the potential of the dime (10,300 V)
b. Calculate the potential 1.00 cm above the dime
(3870 V)