Download Recitation #6b Solution

Physics 212 - Spring 2000
Score:
Recitation Activity 6: Capacitance
NAME: Solution
INSTRUCTOR: Redwing
/100
DATE:_2/11/00
This activity is based on the following concepts:
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A capacitor consists of two isolated conductors with equal and opposite charge +Q and -Q; the charge on the
capacitor is simply referred to as "Q";
The charge Q is proportional to the potential difference V between the two conductors: Q = CV where the
constant C is called the capacitance;
 Capacitance is measured in units of Farads: 1 F = 1 C/V;
Capacitance depends only on the geometry of the capacitor (shape of the conductors, distance between them)
and on the "stuff" (dielectric) you put in between the conductors;
There are 3 principal capacitor geometries that you should become familiar with:
 The parallel plate capacitor
 The cylindrical capacitor
 The spherical capacitor
Suppose a capacitor has a capacitance C when the space between the conducting plates is occupied by vacuum
(or air). Then, the effect of placing a dielectric in this entire space is to increase the capacitance by a factor .
This is known as the "dielectric constant" of the dielectric material.
Capacitors are used to store energy; a capacitor C with a charge Q (or potential difference V between the plates)
stores energy U = Q2/2C = CV2/2. We may view this energy as being stored in the electric field inside the
capacitor.
Capacitors are said to be in PARALLEL when the POTENTIAL DIFFERENCE across their plates is the
SAME; the equivalent capacitance is obtained by adding all the capacitances in parallel.
Capacitors are said to be in SERIES when the CHARGE on their plates is the SAME; the reciprocal of the
equivalent capacitance is obtained by adding the RECIPROCALS of all the capacitances in series.
Questions:
The figure shows a parallel plate capacitor with plates of area A, separated
by a distance d and charged to a potential difference V. The charges on
the plates are +Q and -Q. The space between the plates is occupied by air.
-Q
+Q
Q1. Sketch in the figure the electric field lines between the plates of the
capacitor.
Q2. For the parallel plate capacitor, what equation relates the electric field
E and the potential difference V?
 
V    E.ds  Ed  V  Ed
Q3. For the parallel plate capacitor, what equation relates the electric field
E and the charge Q? (Remember that the plates are conductors and
that the charge Q is uniformly distributed!)
 
  Q
Q


E
 dA   0 EA
E

d
A

0

S
0
d
S
Q4. Write down an expression for the capacitance C in terms of the geometric parameters of the
capacitor?
Q = CV
C = Q/V =
 0 EA
Ed

0 A
d
Analyze the following capacitor circuit:
(Note: a wire is a conductor -- as long as charges do not move in it, all
points on a wire are at the same potential!)
 Capacitors A and C are 1 F each.
 Capacitors B and D are 2 F each.
 The battery shown has a voltage of 9 V across its terminals.
Q5. Identify groups of capacitors (if any) in the circuit that are in
PARALLEL with each other:
A,B are in parallel and C,D are in parallel
A
B
C
9V
Q6. Identify groups of capacitors (if any) in the circuit that are in SERIES
with each other:
Group A,B is in series with group C,D
Q7. Suppose you wanted to replace all the 6 capacitors with a SINGLE capacitor that would
have the same equivalent capacitance. Calculate the value of this single capacitor.
CAB = CA + CB CCD = CC + CD
CAB = 3 F
CCD = 3 F
CSINGLE = (1/ CAB + 1/ CCD)-1
CSINGLE = 1.5 F
Q8. Calculate the charge on each of the 4 capacitors A, B, C, and D.
VAB = VCD = 9 V / 2 = 4.5 V
VA = VB = VAB = 4.5 V VC = VD = VCD = 4.5 V
qA = CAVA = (1 F)(4.5 V) = 4.5 C
qB = CBVB = (2 F)(4.5 V) = 9.0 C
qC = CCVC = (1 F)(4.5 V) = 4.5 C
qD = CDVD = (2 F)(4.5 V) = 9.0 C
Now capacitor A is replaced with a 4 F capacitor.
Q9. What would be the new equivalent capacitance of the circuit?
CAB = CA + CB CCD = CC + CD
CAB = 6 F
CCD = 3 F
CSINGLE = (1/ CAB + 1/ CCD)-1
CSINGLE = 2.0 F
Q10. What is the potential difference across the parallel combination A, B and combination C,
D? Hint: How much charge is stored in combination A, B and combination C, D?
qAB = qCD
CAB VAB = CCD VCD
VAB = CCD VCD / CAB
VAB + VCD = 9 V
CCD VCD / CAB + VCD = 9 V
VCD = 2 * 9 V / 3 = 6 V VAB = 3 V
Q11. Calculate the charge on each of the 4 capacitors A, B, C and D.
VA = VB = VAB = 3.0 V VC = VD = VCD = 6.0 V
qA = CAVA = (4 F)(3.0 V) = 12 C
qB = CBVB = (2 F)(3.0 V) = 6 C
qC = CCVC = (1 F)(6.0 V) = 6 C
qD = CDVD = (2 F)(6.0 V) = 12 C
Scoring Guidelines:
100 All correct or miss one
90
miss two
75
some correct
60
none to few correct, but shows effort
40
Turned in paper, but no real effort
0
No paper
D