Download Electric Potential and Capacitance - Lecture Notes | PHYS 102 and more Study notes Physics in PDF only on Docsity!
Chapter 20: Electric Potential and Capacitance
- The concepts covered in this chapter are:
- Electric potential
- Electric potential energy
- Capacitance
- Capacitors arranged in series and parallel configuration
- Air- and dielectric-filled capacitors
- Potential energy stored in capacitors
- Potential and Potential Energy.
In the previous chapter we learned that charges exert a force on each other according to
Coulomb’s law. This means that such forces that occur among the charges can change a
given configuration of charges: in the field view of electrostatic interaction the electric
fields of various charges interact to make the charges move, i.e., the electric field does
work on the charge configuration). The configuration can also be changed by an external
agent, i.e., work can be performed by an external agent. This change in charge
configuration will show up as a change in the potential energy of the system (the charge
configuration).
The change in the potential energy between (or of ) two charges is defined as
ΔWfield = − ΔWext = − ΔU …[1]
− ΔUab = Ua –Ub =.
b b
a a
∫ F^ dr^ = q^ ∫ E dr^.
JK JJK JJKJJK
…[2]
The potential difference ΔV = Vb - Va is defined as:
Vb - Va =.
b
a
− ∫ E dr
JK JJK
…[3]
Potential can be defined as the potential energy per unit charge ( recall a similar
definition of the electric field E as the force per unit charge, q E = F )
A point charge Q, creates an electric potential V(r) at a distance r from it given by:
V(r) = kQ/r ( referenced to infinity, i.e., V ( ∞ ) = 0 ) …[4]
The net potential at a given point due to a collection of charges is simply the algebraic
sum of individual potentials
1
n i i (^) i
Q
V k = r
= ∑ …[5]
Potential energy of two point charges is given by
U 12 = kQ 1 Q 2 /r …[6]
Potential energy of a collection or configuration of point charges is the algebraic sum
of the potential energy of all the pairs of charges you can form in the configuration. For
example, in Solved Example 1 below, Ui is the potential energy of the 3-charge
configuration.
Potential energy of a single charge in a collection or configuration of point charges is
the algebraic sum of the potential energy of all the pairs the given charge forms in the
configuration. For example, Solved Example 1 the potential energy of Q 1 in part [a]
would be the sum U 12 + U 13.
Points to remember :
- Make sure you understand the difference between potential and potential energy, and
between potential energy of a charge configuration and the potential energy of a charge in
a charge configuration.
- A single charge will create a potential at a point in space, but you need at least two
charges for the potential energy to have any meaning.
- Potential and potential energy are scalars.
- Eq. [3] can also be expressed as :
4. Solved Examples. 1. Three charges are arranged at the vertices of a right-angle triangle as shown, with Q 1 = 4 μ C =2Q 2 , and Q 3 = - 3 μ C.
53 o
5.0cm
Q (^1)
Q 3 Q^2 X
Y
[a] Determine the total potential energy of the three-charge system.
[b] Determine the work done by an external force to move Q 2 from its present position to infinity. [a] The initial potential energy, 12 12 12 9 2 2 2
i 3 10 5 10 4 10
U J
− − − − − −
− × × ×
= × − + = −
× × ×
[b] after Q 2 has been removed to infinity, the potential energy of the remaining charges is 12 9 2
U (^) f J
− −
− ×
= × = −
×
the work done by the external force is given by Δ W = Δ U = Uf - Ui = - 3.60 – ( - 2.88) =
- 0.72J
Note : in the problem above we have used Δ W = Δ U = Uf - Ui , where Δ U is the change in
the potential energy of the charge configuration. We could have also done the problem by looking at it from the point of view of Q 2 (in a manner of speaking, charges don’t
really have a point of view). In this case Δ U would be the change in the potential energy
of Q 2.
Δ U = Uf(Q 2 ) - Ui(Q 2 ) = 0 – (U 12 + U 23 ) = - 9*10^9 (- 1.2 +2.0)10-10^ = - 0.72J
2. (Prob. 20.18 from the text) discussed in lecture This is a rather longish method but has several explanatory details. The problem can also be solved in a much more compact way. The shorter method follows the long one. (a) In an empty universe, the 20-nC charge can be placed at its location with no energy investment. At a distance of 4 cm, it creates a potential
V
1 (^9 2 2 )(^9 )
1
9 10 N m C 20 10 C 4.50 kV 0.04 m
k q e r
× ⋅ ×^ − = = =
To place the 10-nC charge there we must put in energy
U 12 = q 2 V 1 = ( 10 × 10 −^9 C)( 4.5 × 10 3 V)= 4.50 × 10 −^5 J.
Next, to bring up the –20-nC charge requires energy ( )
23 13 3 2 3 1 3 2 1 9 9 2 2 9 9
5 5
10 10 C 20 10 C 20 10 C 9 10 N m C 0.04 m 0.08 m 4.50 10 J 4.50 10 J
U U q V q V q V V −^ −^ −
− −
- = + = + ⎛ × × ⎞ = − × × ⋅ (^) ⎜ + ⎟ ⎝ ⎠ = − × − ×
The total energy of the three charges is
U 12 + U 23 + U 13 = −4.50 × 10 −^5 J.
(b) The three fixed charges create this potential at the location where the fourth is released:
9 2 2 9 9 9 (^1 2 3 2 ) 3
20 10 10 10 20 10 9 10 N m C C m 0.04 0.03 0.03^ 0. 3.00 10 V
V V V V
V
⎛ × −^ × −^ × − ⎞ = + + = × ⋅ (^) ⎜ + − ⎟ ⎝ (^) + ⎠ = × Energy of the system of four charged objects is conserved as the fourth charge flies away: 1 2
⎛⎝⎜ mv (^2) + qV ⎞⎠⎟ i
= 1 2
⎛⎝⎜ mv (^2) + qV ⎞⎠⎟ f
0 + ( 40 × 10 −^9 C)( 3.00 × 10 3 V)=
1 2
(2.00 ×^10 −^13 kg) v^^2 +^0
v =
2 1.20( × 10 −^4 J)
2 × 10 −^13 kg
= 3.46 × 10 4 m s
The shorter method.
[a] U = U 12 + U 23 + U 13 = U 13 = - 910^9 40010-18/810-2^ = - 4.5* 10-5^ J
(why is U 12 + U 23 = 0 ?)
[b] (U + KE)i = (U + KE)f ( conservation of energy from last term)
[b] Q 4 = Vab 4*10-6^ = 8*10-5^ C
[c] U = ½( 6 *10-6)( Vab)^2 = 1.2*10-3J
4. (prob 41 solved in lecture)
(a) 1 Cs
= 1
Cs = 2.50 μ F C (^) p = 2.50 + 6.00 = 8.50 μ F
Ceq = 1 8.50 μ F
⎛ ⎝⎜^
⎞ ⎠⎟
− 1 = 5.96 μ F
(b) Q^ =^ C Δ V^ =^ (5.96 μ^ F) (15.0 V ) =^ 89.5^ μ^ C^ on 20.0 μ F
Δ V = Q C
= 89.5^ μ^ C 20.0 μ F
= 4.47 V
15.0 − 4.47 = 10.53 V
Q = C Δ V = (6.00 μ F) (10.53 V ) = 63.2 μ C on 6.00 μ F
89.5 − 63.2 = 26.3 μ C on 15.0 μ F and 3.00 μ F
Concept Questions
- A spherical balloon contains a positively charged object at its center. (i) As the balloon is inflated to a greater volume while the charged object remains at the center, does the electric potential at the surface of the balloon (a) increase, (b) decrease, or (c) remain the same? (ii) Does the electric flux through the surface of the balloon (a) increase, (b) decrease, or (c) remain the same?
Answer: (i), (b). The electric potential is inversely proportional to the radius (see Eq. 20.11). (ii), (c). Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant.
r
q V = ke [Eq. 20.11]
- In Active Figure 20.6a, take q 1 to be a negative source charge and q 2 to be the test charge. (i) If q 2 is initially positive and is replaced with a charge of the same magnitude but negative, does the potential at the position of q 2 due to q 1 (a) increase, (b) decrease, or (c) remain the same? (ii) When q 2 is changed from positive to negative, does the potential energy of the two-charge system (a) increase, (b) decrease, or (c) remain the same?
Active Figure 20.6 (a) If two point charges are separated by a distance r 12 , the potential energy of the pair of charges is given by keq 1 q 2 / r 12. (b) If charge q 1 is removed, a potential keq 2 / r 12 exists at point P due to charge q 2.
- You have three capacitors and a battery. In which of the following combinations of the three capacitors will the maximum possible energy be stored when the combination is attached to the battery? (a) When in series the maximum amount is stored. (b) When parallel the maximum amount is stored. (c) Both combinations will store the same amount of energy.
Answer: (b). For a given voltage, the energy stored in a capacitor is proportional to C : U = C (Δ V )^2 /2. Therefore, you want to maximize the equivalent capacitance. You do so by connecting the three capacitors in parallel so that the capacitances add.
