College Physics 150 Chapter 17 – Electric Potential
• Electric Potential Energy • Electric Potential • How are the E-‐‑field and Electric Potential related? • Motion of Point Charges in an E-‐‑field • Capacitors • Dielectrics
Electric Potential Energy Electric potential energy (Ue) is energy stored in the electric field.
• Ue depends only on the location, not upon the path taken to get there (conservative force). • Ue = 0 at some reference point. • For two point particles take Ue = 0 at r = ∞. • For the electric force
kq1q2 Ue = r
Example: A proton and an electron, initially separated by a distance r, are brought closer together. (a) How does the potential energy of this system of charges charge? 2 ke For these two charges U = − e r
Bringing the charges closer together decreases r: ΔU
e
= U ef − U ei < 0
This is like a mass falling near the surface of the Earth; positive work is done by the field. (b) How will the electric potential energy change if both particles have positive (or negative) charges? When q1 and q2 have the same algebraic sign then ΔUe > 0. This means that work must be done by an external agent to bring the charges closer together.
What is the potential energy of a system (arrangement) of point charges? To calculate: Begin by placing the first charge at a place in space far from any other charges. No work is required to do this. Next, bring in the remaining charges one at a time until the desired configuration is finished. Example: What is the potential energy of three point charges arranged as a right triangle? (See text Example 17.2)
q2
q2
r12
q1
r12
r23
r13
q3
kq1q2 kq1q3 kq2 q3 Ue = 0 + + + r12 r13 r23
q1
r13
r23
q3
kq1q2 kq1q3 kq2 q3 + + Ue = 0 + r13 r23 r12
Are these the same?
Electric Potential
Electric potential is the electric potential energy per unit charge.
Ue V= qtest Electric potential (or just potential) is a measurable scalar quantity. Its unit is the volt (1 V = 1 J/C).
Ue kQ = For a point charge of charge Q: V = qtest r When a charge q moves through a potential difference of ΔV, its potential energy change is ΔUe = qΔV.
Example: A charge Q = +1 nC is placed somewhere in space far from other charges. Take ra = rb = rc = rd = 1.0 m and re = rf = rg = 2.0 m. f b
e
a
Q
d
c
(a) Compare the potential at points d and g. Since Q > 0 the potential at point d is greater than at point g, it is closer to the charge Q.
g
(b) Compare the potential at points a and b. The potential at point a is the same as at point b; both are at the same distance from the charge Q.
Example: A charge Q = +1 nC is placed somewhere in space far from other charges. Take ra = rb = rc = rd = 1.0 m and re = rf = rg = 2.0 m. f b
e
a
c
Q
d
g
(c) Place a charge of +0.50 nC at point e. What will the change in potential (ΔV) be if this charge is moved to point a?
Ve
kQ ( 9.0 ×10 = =
Va
kQ ( 9.0 ×10 = =
9
re
ra
Nm 2 /C 2 ) (1.0 nC) 2m
9
Nm 2 /C 2 ) (1.0 nC) 1m
= +4.5 Volts = +9.0 Volts
ΔV = Vf -‐‑ Vi = Va -‐‑ Ve = +4.5 Volts (d) What is the change in potential energy (ΔU) of the +0.50 nC charge ? ΔUe = qΔV = (+0.50 nC)(+4.5 Volts)= +2.3 nJ
Example: A charge Q = +1 nC is placed somewhere in space far from other charges. Take ra = rb = rc = rd = 1.0 m and re = rf = rg = 2.0 m. f b
e
a
c
Q
d
(e) How would the results of the previous questions change if instead of a +1.0 nC charge there is a -‐‑1.0 nC charge in its place?
g
(a) The potential at point d is less than the potential at point g. (b) Unchanged (c) -4.5 V (d) -2.3 nJ
The Relationship between E and V The circles are called equipotentials (surfaces of equal potential).
f
f b
e
a
Q
d
b
c e
+9 V
a
+4.5 V
g
The electric field will point in the direction of maximum potential decrease and will also be perpendicular to the equipotential surfaces.
Q
d
g
c +9 V
+4.5 V
Equipotentials and field lines for a dipole.
Uniform E-‐‑field
V1
V2
V3
V4 E
Equipotential surfaces
ΔU e ΔV = = − Ed q
Where d is the distance over which ΔV occurs.
If the electric field inside a conductor is zero, what is the value of the potential? If E = 0, then ΔV = 0. The potential is constant! What is the value of V inside the conductor? It will be the value of V on the surface of the conductor.
Moving Charges When only electric forces act on a charge, its total mechanical energy will be conserved.
Ei = E f Example (text problem 17.40): Point P is at a potential of 500.0 kV and point S is at a potential of 200.0 kV. The space between these points is evacuated. When a charge of +2e moves from P to S, by how much does its kinetic energy change?
Ei = E f K i +Ui = K f +U f
K f − K i = Ui −U f = − (U f −Ui ) = −ΔU = −qΔV = −q (Vs −Vp ) = − (+2e) ( 200.0 − 500.0 ) kV = +9.6 ×10 −14 J
Example (text problem 17.41): An electron is accelerated from rest through a potential difference. If the electron reaches a speed of 7.26×106 m/s, what is the potential difference?
Ei = E f 0
K i +Ui = K f +U f
K f = −ΔU = −qΔV 1 2 mv f = −qΔV 2 9.11×10 kg) ( 7.26 ×10 m/s) mv ( ΔV = − =− 2q 2 (−1.60 ×10 −19 C) 2 f
= +150 Volts
−31
6
Note: the electron moves from low V to high V.
2
Capacitors A capacitor is a device that stores electric potential energy by storing separated positive and negative charges. Work must be done to separate the charges.
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+
+
+
+
+
+
–
–
–
–
–
–
–
For a parallel plate capacitor:
Parallel plate capacitor
E ∝Q E ∝ ΔV ∴Q ∝ ΔV
Wrijen as an equality: Q = CΔV, where the proportionality constant C is called the capacitance.
Capacitors A capacitor is a device that stores electric potential energy by storing separated positive and negative charges. Work must be done to separate the charges.
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+
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–
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–
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What is the capacitance for a parallel plate capacitor?
Parallel plate capacitor
σ Q ΔV = Ed = d = d ε0 ε0 A ε0 A ∴Q = ΔV = CΔV d ε0 A where C = . d
Note: C depends only on constants and geometrical factors. The unit of capacitance is the farad (F). 1 F = 1 C2/J = 1 C/V
Example (text problem 17.56): A parallel plate capacitor has a capacitance of 1.20 nF. There is a charge of magnitude 0.800 µC on each plate. (a) What is the potential difference between the plates?
Q = CΔV Q 0.800 µC ΔV = = = 667 Volts C 1.20 nF (b) If the plate separation is doubled, while the charge is kept constant, what will happen to the potential difference?
Q Qd ΔV = = C ε0 A ΔV ∝ d If d is doubled so is the potential difference.
Example (text problem 17.100): A parallel plate capacitor has a charge of 0.020 µC on each plate with a potential difference of 240 volts. The parallel plates are separated by 0.40 mm of air. (a) What is the capacitance of this capacitor?
Q 0.020 µC C= = = 8.3×10 −11 F = 83 pF ΔV 240 Volts (b) What is the area of a single plate?
ε0 A C= d Cd (83 pF ) ( 0.40 mm ) A= = ε 0 8.85 ×10 −12 C 2 /Nm 2 = 0.0038 m 2 = 38 cm 2
Dielectrics As more and more charge is placed on capacitor plates, there will come a point when the E-‐‑field becomes strong enough to begin to break down the material (medium) between the capacitor plates. To increase the capacitance, a dielectric can be placed between the capacitor plates.
C = κ C0
ε0 A where C0 = d
and κ is the dielectric constant.
Example (text problem 17.71): A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are 0.3 m by 0.4 m and the waxed paper, of slightly larger dimensions, is of thickness 0.030 mm and has κ = 2.5, what is the capacitance of this capacitor?
ε0 A C0 = d −12 2 2 2 8.85 ×10 Nm /C 0.40 * 0.30 m ( ) ( ) = 0.030 ×10 −3m = 3.54 ×10 −8 F and C = κ C 0 = ( 2.5) (3.54 ×10 −8 F ) = 8.85 ×10 −8 F.
Energy Stored in a Capacitor A capacitor will store energy equivalent to the amount of work that it takes to separate the charges.
The energy stored in the electric field between the plates is:
1 U = QΔV 2 1 2 = C ( ΔV ) 2 Q2 = 2C
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These are found by using Q = CΔV and the first relationship.
Example (text problem 17.79): A parallel plate capacitor is composed of two square plates, 10.0 cm on a side, separated by an air gap of 0.75 mm.
(a) What is the charge on this capacitor when the potential difference is 150 volts?
ε0 A Q = CΔV = ΔV = 1.77 ×10 −8 C d (b) What energy is stored in this capacitor?