Hour Exam #1
September 20, 1999
Possibly useful information:
k = 9 x 109 N m2/C2e = 1.6 x 10 - 19 C
Ceq = C1 + C2 + C3
P = IV = I2R = V2/R
Req = R1 + R2 + R3 + . . .
1. What is the electric field at the origin (x = 0, y = 0) when charges Q1 = - 10 µC and Q2 = - 30 µC are placed at corners of a square, 10.0 cm (0.10 m) on a side, as shown in the figure below?
What is the electric field -- magnitude and direction -- at the origin?
This is almost homework problem 17.18.
First we will calculate E1, the field due to charge Q1 and find its direction. Then we will calculate E2, the field due to charge Q2 and find its direction. Then we will find the vector sum of these two electric fields for the net field.
E1 = 9.0 x 106 N/C
and this field points upward. That is,
E1x = 0 E1y = + 9.0 x 106 N/C
Now for the field due to the other charge, Q2,
E2 = 2.7 x 107 N/C
and this field points to the right. That is,
E2x = 2.7 x 107 N/C E2y = 0
The net field is the vector sum of these,
Enet = E1 + E2 which is shorthand notation for
Enet,x = E1x + E2x and
Enet,x = 0 + 2.7 x 107 N/C
Enet,x = + 2.7 x 107 N/C
Enet,x = + 27 x 106 N/C
Now we have the components of the field and we can construct the field itself -- that is, we can calculate its magnitude and direction.
E = 28.5 x 106 N/C
= 18.4°
That is, the net electric field is
E = 28.5 x 106 N/C at the angle as shown in the diagram.
2. What is the electric potential at the origin (x = 0, y = 0) due to charges Q1 = - 10 µC located at (x = 0, y = 0.10 m) and Q2 = - 30 µC located at (x = 0.10 m, y = 0). The electric potential is zero infinitly far from the charges.
The electric potential due to a point charge is
Vtot =
Vi = V1 + V2
V1 = k Q1 / r1
V1 = ( 9 x 109 ) ( - 10 x 10 - 6 ) / 0.10
V1 = - 9 x 105 volts
V2 = k Q2 / r2
V2 = ( 9 x 109 ) ( - 30 x 10 - 6 ) / 0.10
V2 = - 2.7 x 106 volts = - 27 x 105 volts
Vtot = V1 + V2
Vtot = ( - 9 - 27 ) x 105 volts
Vtot = - 36 x 105 volts
Vtot = - 3.6 x 106 volts
3. Three capacitors are connected in the circuit shown below. They have values of C1 = 100
F, C2 = 200
Find the charge on each capacitor and the potential energy stored in capacitor C1.
First, we replace the parallel capacitors, C2 and C3, with a single equivalent capacitor which we will call Ceq1:
Ceq1 = C2 + C3
Ceq1 = (200 + 300)
Ceq1 = 500
Now we replace these series capacitors, C1 and Ceq1, with a single equivalent capacitor which we will call Ceq:
Ceq = 83.3
C = Q/V ===> Q = C V
Q = (83.3 x 10 - 6 F)(12 V)
Q = 1 x 10 - 3 C
PETot = [1/2] Q V = [1/2] [1 x 10 - 3 C] [12 V]
PETot = 6 x 10 - 3 J = 6 mJ
Now we return to the individual capacitors,
Q1 = Q = 1 x 10 - 3 C
Note that the voltage across capacitor C1 is not 12 V !
V1 = 10.0 V This means the voltage across the parallel part, C2 and C3, must be
V2 = V3 = 2.0 V PE1 = [1/2] Q1 V1 = 5 x 10 - 3 J = 5 mJ
V2 = V3 = 2.0 V
Note that V2 is not 12.0 V !
Q2 = C2 V2 = (200 x 10 - 6 F) (2.0 V) = 4 x 10 - 4 C = Q2 Q2 = 4 x 10 - 4 C
PE2 = [1/2] Q2 V2 = [1/2] ( 4 x 10 - 4 C ) (2 V)
PE2 = 4 x 10 - 4 J = 0.4 mJ
V3 = 2.0 V
Q3 = C3 V3
Q3 = (300 x 10 - 6 F) (2.0 V) = 6 x 10 - 4 C
PE3 = [1/2] Q3 V3 = [1/2] ( 6 x 10 - 4 C ) (2 V)
PE3 = 6 x 10 - 4 J = 0.6 mJ
Notice that
PETot = 6 mJ = 5 mJ + 0.4 mJ + 0.6 mJ = PE1 + PE2 + PE3 and that
Q = Q1 = 1 x 10 - 3 C = 4 x 10 - 4 C + 6 x 10 - 4 C = Q2 + Q3
4. Find the equivalent resistance in the circuit shown below for R1 = 100
, R2 = 500
Resistors R2 and R3 are in parallel so we begin by replacing them with an equivalent resistor Req whose value we can calculate from
1/Req = 1/R2 + 1/R3 = 1/500 1/Req = (0.003) /
Req = [1/0.003]
Now we have R1, this Req, and R4 in series as in the following diagram,
or
The new equivalent resistor Req,2 for this combination of three resistors in series is given by
Req,2 = R1 + Req + R4 Req,2 = 100
Req,2 = 483
The current from the battery is given by
I = V / Req,2 = 12 V / 483 This is the current through R1 and R4,
I1 = I4 = 0.0248 A The power supplied from the battery is
P = I V = (0.0248 A)(12 V) = 0.2981 W = 0.30 W Pbat = 0.30 W
Now we work back and find the power absorbed by each of the resistors. We know the current through R1 and R4 so we can find the voltage across them and then find the power absorbed by them.
V1 = I R1 = (0.0248 A)(100 P1 = I V1 = (0.0248 A)(2.48 V) = 0.0615 W
V4 = I R4 = (0.0248 A)(50
P4 = I V4 = (0.0248 A)(1.24 V) = 0.0307 W
The voltage across the resistors in parallel -- R2 and R3 -- is the 12 V of the battery minus the voltage across R1 and R4,
V2 = V3 = 12 V - 2.48 V - 1.24 V = 8.28 V P = I V = (V / R) V = V2/R
P2 = (8.28 V)2 / (500
P3 = (8.28 V)2 / (1000
I2 = V2/R2 = 8.28 V / 500
I3 = V3/R3 = 8.28 V / 1000
As a quick check, see if these absorbed powers equal the battery's output power,
Ptot = P1 + P2 + P3 + P4 Ptot = (0.0615 + 0.069 + 0.137 + 0.0307) W
Ptot = 0.2982 W = 0.2891 W = Pbat
And that agrees to three significant figures.
Now calculate the currents through R2 and R3,
I2 = V2 / R2 = 8.28 V / 500 I3 = V3 / R3 = 8.28 V / 1000
Notice, too, that
I1 = I4 = I2 + I3 = 0.0248 A
5. a) What is the final charge Qf that will accumulate on the capacitor in this circuit, after switch S is closed, if we wait long enough?
b) What is the time constant of this circuit?
c) What charge will there be on the capacitor 1.0 s after the switch is closed?
After a long time, the current will be zero because the voltage across the capacitor will just "balance" the voltage of the battery,
Vf = 12 v C = Q / V
Qf = C Vf
Qf = ( 500 x 10 - 6 f ) ( 12 v )
Qf = 6.0 x 10 - 3 C
The time constant of this circuit is
= RC
The charge on the capacitor, as a function of time, is
q = q(t) = Qf [ 1 - e - t / q ( 1.0 s) = q (t = 1.0 s) = Qf [ 1 - e - 1.0 /
q ( 1.0 s) = Qf [ 1 - e - 1.0 / 0.5 ]
q ( 1.0 s) = Qf [ 1 - ( e - 2 ) ]
q ( 1.0 s) = Qf [ 1 - ( 0.135 ) ]
q ( 1.0 s) = Qf [ 0.865 ]
q ( 1.0 s) = 0.865 Qf
q ( 1.0 s) = 0.865 ( 6.0 x 10 - 3 C )
q ( 1.0 s) = 5.19 x 10 - 3 C
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