Homework 3
1.
A potential difference of 128.0 V is applied across the two collinear conducting cylinders shown in the figure. The radius of the outer cylinder is 15.0
cm, the radius of the inner cylinder is 8.0 cm, and the length of the two cylinders is 39.0 cm. What is the magnitude of the electric field just outside the
surface of the inner cylinder (r1)?
2.545×10
3
V/m
Potential difference: V=q/(Eo*2*pi*L) * ln(r2/r1), solve for q, then E(N/C)=q/(Eo*2*pi*r1*L)
2.
A potential difference of
V
= 32.0 V is applied across a circuit with capacitances
C
1
= 1.3 nF,
C
2
= 4.1 nF, and
C
3
= 3.7 nF, as shown in the figure.
a) What is the magnitude and sign of
q
3l
, the charge on the left plate of
C
3
(marked by point
A
)?
7.026×10
8
C
C1+C2 in parallel, C3 in series. 1/C123=C12*C3/(C12+C3)
b) What is the electric potential difference,
V
3
, across
C
3
?
19.0 V
Electric potential difference: V=q/C3
c) What is the magnitude and sign of the charge
q
2r
, on the right plate of
C
2
(marked by point
B
)?
5.335×10
8
C
Magnitude of q2: (C2/(C1+C2))*Answer a
3.
Six capacitors are connected as shown in the figure (Pg 131).
a) If the capacitance of capacitor 3 is 2.5 nF, what does
C
2
have to be to yield an equivalent capacitance of 5.2 nF for the combination of capacitors 2 and 3?
2.700 n
b) For the same values of
C
2
and
C
3
as in part (a), what is the value of
C
1
that will give an equivalent capacitance of 2.229 nF for the combination of the three capacitors?
3.900 nF
c) For the same values of the capacitances of capacitors 1, 2 and 3 as in part (b), what is the equivalent capacitance (in nF) of the whole system of capacitors, if the values of the remaining
are
C
4
= 1.3 nF,
C
5
= 4.9 nF, and
C
6
= 2.7 nF? (Series)
1.782 nF
C123*C456/(C123+C456)
d) If a battery with a potential difference of 9.1 V is connected to the capacitors as shown in the figure, what is the total charge on the six capacitors?
16.219 nC
q=V*Ceq
e) What is the potential drop across capacitor 5 in this case?
1.822 V
V=q/C456
4.
The potential difference across two capacitors in series is 290. V. The capacitances are
C
1
= 1.17 mF and
C
2
= 1.941 mF.
a) What is the total capacitance of this pair of capacitors?
0.730 mF
Series: 1/C12=1/C1+1/C2
b) What is the charge on each capacitor?
0.2117 C
q=VCeq
c) What is the potential difference across capacitor 1?
180.9 V
V=q/C1
d) What is the total energy stored by the capacitors?
3.070×10
1
J
E=½*q*290(V)
5.
A 1780.nF(10^9F) parallel plate capacitor is connected to a 11.6V battery and charged.
a) What is the charge
Q
on the positive plate of the capacitor?
2.065×10
5
C
q=V*C
b) What is the electric potential energy stored in the capacitor?
1.198×10
4
J
U=1/2*q*V
6.
A 3.1nF parallel plate capacitor with a sheet of Mylar (κ = 3.1) filling the space between the plates is charged to a potential difference of 289.0 V and
is then disconnected.
a) How much work is required to completely remove the sheet of Mylar from the space between the two plates?
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 Spring '07
 Gade
 Electric Potential, Energy, Potential difference, coulombs, vEMF

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