Quiz 5 Solutions
Question
A 10 F capacitor is charged with a battery of 10V. After charging, this capacitor is disconnected from the battery and connected to another capacitor of 5 F.
a. Find the initial charge on the 10 F capacitor and the potential energy stored in it.
b. Find the final potential difference across the capacitors, and the charges on each of
the capacitors.
c. Find the final potential energy stored in the two capacitors.
d. How much is the total final potential energy? Is it same or different than the initial
potential energy?
e. (Bonus question, no penalty if wrong) Depending on what you find, explain why is
the total final potential energy same/different than the initial potential energy?
Solution
a) Voltage across the 10µF is 10V. Therefore the charge on it Qintial = CV = 10 × 10 =
100µC, and energy is U = CV 2 /2 = 5 × 10( − 4)J. Let the charge on this be Qinitial .
b) After charging the 10µF (call it capacitor 1) is connected to uncharged 5µF (capacitor
2). Charge will thus flow from the charged capacitor to the uncharged. In this flow of charge
though, the total initial charge will be conserved. Thus,
Qinitial = Q1,f inal + Q2,f inal
(1)
The charges flow because there is an electric force pushing them. This electric field is
coming from the difference in potentials of the two capacitors. Capacitor 1 starts from 10V,
and capacitor 2 starts from 0V. As charge is transferred from capacitor 1, Q1 decreases
and Q2 increases, and because V = Q/C, voltage on capacitor 1 decreases, and voltage on
capacitor 2 increases. Thus, the charge is transferred until the two capacitors have the same
potentials, V1,f inal = V2,f inal , which using V = Q/C gives
Q1,f inal
Q2,f inal
=
= V2,f inal
C1
C2
Plugging values for C1 and C2 , gives
V1,f inal =
Q1,f inal = 2Q2,f inal
1
(2)
(3)
Solving eqs. (1), (3) we get, Q1,f inal = Qinitial × 2/3 = 66.67µC, and Q2,f inal =
Qinitial /3 = 33.33µC, and V1,f inal = Q1,f inal /C1 = 6.67V, which is the same as V2,f inal .
Since there is no battery in the circuit, if you draw the circuit diagram of the capacitor
1 connected to capacitor 2, it may seem that they are in series or in parallel. Since the
formulas for capacitors in series or parallel are different, we need to know which is it - series
or parallel? The answer is parallel. Why? To be in series capacitors must have the same
charge, and to be in parallel the capacitors must have same potential. We saw that the final
potentials on both the capacitors should be equal, and thus the capacitors are in parallel.
Although the same current flows through each capacitor (indicating that they could be
in series), the capacitors have different charges from each other since their initial charges
were different. Also, the current takes charge from capacitor 1 to capacitor 2, and thus
it decreases Q1 , but increases Q2 . In the case of a battery connected to two capacitors
in series, the current charges both the capacitors equally, and additionally we assume that
the capacitors are uncharged before - thus the two capacitors have the same charge as the
battery charges both of them. If the two capacitors had different initial charges before you
connected the battery with them, the series formulas will not work!
c) Since we know the charges and potentials on both of the capacitors, we can calculate
−4
2
2
the potential energies easily. U1 = C1 V1,f
inal /2 = 2.22 × 10 J, and U2 = C2 V2,f inal /2 =
1.11 × 10−4 J.
d) The total final potential energy is Uf inal = U1 + U2 = 3.33 × 10−4 J, and is l ess than
Uinitial = 5 × 10−4 J.
e) Since Uf inal < Uinitial , some energy has been lost - to what? Our first guess would be
heat, but for that you need resistance, and we have none in our problem! We assumed ideal
wires, but in reality they would have some small resistance R, and we shall see that it is this
small resistance which leads to energy loss - in fact the resistance of the wires does not enter
the equation at all, and the energy lost is independent of this R.
Figure 1: Circuit diagram with resistance of wires taken into account as the resistor R
We model the resistance of the wires in the circuit as a small resistance R connected to
the capacitors (see figure above). To find the current through the circuit we would have to
2
solve the Kirchoff’s loop rule which states
V1 (t) = V2 (t) + VR (t)
(4)
where V1 , V2 are the voltages on capacitors 1 and 2, and VR is the voltage across the small
resistance R of the wires. All these depend on time as the current varies in time - it starts
at non-zero value as charge flows from capacitor 1 to 2, and finally becomes zero after the
capacitors reach the same voltage. In terms of this time-dependent current i(t), the voltages
can be expressed as
Z t
1
Q1 (t)
(5)
i(t0 )dt0
=
Qinitial −
V1 (t) =
C1
C1
0
Q2 (t)
1 Zt 0 0
V2 (t) =
=
i(t )dt
C2
C2 0
(6)
VR (t) = i(t)R
(7)
Using these equations, we get a differential equation for i(t) (after differentiation with t)
1
1 1
di(t)
+
i(t)
=−
dt
R C1 C2
(8)
which we can solve easily to get
t
i(t) = i0 exp −
RCeq
!
(9)
where Ceq = (1/C1 + 1/C2 )−1 is the equivalent capacitance of C1 and C2 in series. i0 is the
current at t = 0, which can be fixed by noting that V1 (t = 0) = 10V and V2 (t = 0) = 0V ,
and using eq. (4) VR (t = 0) = 10V which gives current i0 = VR /R = 10/R. As expected the
current decays to 0 as t → ∞, as the capacitors reach the same potentials.
Since we know the current in the circuit we can find power dissipated by the resistor
P (t) = i2 (t)R which we can integrate to find the total energy emitted by the resistor
Z ∞
Z ∞
Z ∞"
10
P (t)dt =
i (t)Rdt =
t
exp −
R
RCeq
!#2
100
(2RCeq ) = 200Ceq
R
0
0
0
(10)
Notice that R drops out of the energy dissipated, and thus the energy dissipated by the
resistor R is independent of R! Using Ceq = 10/3µF , we find that Udissipated = 1.67 × 10−4 J
which is exactly the difference of Uinitial and Uf inal found before!
Udissipated =
2
3
Rdt =