**PHY1160C
Exam #1
September 15, 2000
Name ______________________________**

This exam is fairlylengthyorlong. Pace yourself accordingly. Dospend too much time on any one problem.not

It will be easy to spend too much time on the first two problems. Resist that temptation. Spread your time and efforts overof the questions.all

Possibly useful information:

k = 9 x 10

^{9}N m^{2}/C^{2}e = 1.6 x 10

^{ - 19}C

m_{e}= 9.11 x 10^{ - 31}kg

PE = k Q q / r

V = k Q / r

_{}C

_{eq}= C_{1}+ C_{2}+ C_{3}

_{}

P = IV = I^{2}R = V^{2}/R

_{}

R_{eq}= R_{1}+ R_{2}+ R_{3}+ . . .

q = Q_{f}e^{ - t/RC}i = I

_{o}e^{ - t/RC}

q = Q_{f}[ 1 - e^{ - t/RC}]i = I

_{o}[ 1 - e^{ - t/RC}]

**1. ** What is the electric field at
the origin (x = 0, y = 0) when charges
**Q _{1} = - 10 µC** and

Diagramshelp,always

First we will calculate
**E _{1}**, the field due to charge Q

E _{1}= 9.0 x 10^{6}N/Cand this field points

upward. That is,E _{1x}= 0E

_{1y}= + 9.0 x 10^{6}N/CNow for the field due to the other charge, Q

_{2},

E _{2}= 2.7 x 10^{7}N/Cand this field points to the

right. That is,E _{2x}= 2.7 x 10^{7}N/CE

_{2y}= 0The net field is the

vectorsum of these,E_{net}= E_{1}+ E_{2}which is shorthand notation for

E _{net,x}= E_{1x}+ E_{2x}and E_{net,y}= E_{1y}+ E_{2y}E

_{net,x}= 0 + 2.7 x 10^{7}N/C E_{net,y}= + 9.0 x 10^{6}N/C + 0E

_{net,x}= + 2.7 x 10^{7}N/C E_{net,y}= + 9.0 x 10^{6}N/CE

_{net,x}= + 27 x 10^{6}N/C E_{net,y}= + 9.0 x 10^{6}N/C

E = 28.5 x 10 ^{6}
N/C

= 18.4°

That is, the net electric fieldEisE = 28.5 x 10 ^{6}N/C at the angle = 18.4^{o}as shown in the diagram.

The electric potential due to a point charge isV

_{tot}=_{}V_{i}= V_{1}+ V_{2}V

_{1}= k Q_{1}/ r_{1}V

_{1}= ( 9 x 10^{9}) ( - 10 x 10^{ - 6}) / 0.10V

_{1}= - 9 x 10^{5}voltsV

_{2}= k Q_{2}/ r_{2}V

_{2}= ( 9 x 10^{9}) ( - 30 x 10^{ - 6}) / 0.10V

_{2}= - 2.7 x 10^{6}volts = - 27 x 10^{5}voltsV

_{tot}= V_{1}+ V_{2}V

_{tot}= ( - 9 - 27 ) x 10^{5}voltsV

_{tot}= - 36 x 10^{5}volts

C _{eq1}= C_{2}+ C_{3}C

= (250 + 500) F_{eq1}C

_{eq1}= 750 F

C

_{eq}= 125 FC = Q/V ===> Q = C V

Q = ( 125 x 10

^{ - 6}F )( 12 V )Q = 1.5 x 10

^{ - 3}CNow, for the total energy supplied by the battery,

PE _{Tot}= [^{1}/_{2}] Q V = [^{1}/_{2}] [1.5 x 10^{ - 3}C] [12 V]PE

_{Tot}= 9 x 10^{ - 3}J = 9 mJ

Now we return to the

individualcapacitors,Q_{1}= Q = 1.5 x 10^{ - 3}CNote that the voltage across capacitor C

_{1}isnot12 V !V _{1}= 10.0 VThis means the voltage across the parallel part, C

_{2}and C_{3}, must beV _{2}= V_{3}= 12 V - V_{1}= 12 V - 10 VV

_{2}= V_{3}= 2.0 VNow, for the potentential energy stored in the capacitors,

PE _{cap}= [^{1}/_{2}] Q VPE

_{1}= [^{1}/_{2}] Q_{1}V_{1}PE

_{1}= [^{1}/_{2}] ( 1.5 x 10^{ - 3}C )( 10 V ) = 7.5 x 10^{&endash; 3}J = 7.5 mJ

PE_{1}= 7.5 mJNow for the other capacitors,

V _{2}= V_{3}= 2.0 VNote that V

_{2}isnot12.0 V!Q _{2}= C_{2}V_{2}= (250 x 10^{ - 6}F) (2.0 V) = 5 x 10^{ - 4}C = Q_{2}Q

_{2}= 5 x 10^{ - 4}C = 0.5 x 10^{ - 3}C

Q_{2}= 0.5 x 10^{ - 3}CPE

_{2}= [^{1}/_{2}] Q_{2}V_{2}= [^{1}/_{2}] ( 5 x 10^{ - 4}C ) (2 V)PE

_{2}= 5 x 10^{ - 4}J = 0.5 mJV

_{3}= 2.0 VQ

_{3}= C_{3}V_{3}Q

_{3}= (500 x 10^{ - 6}F) (2.0 V) = 1 x 10^{ - 3}C

Q_{3}= 1 x 10^{ - 3}CPE

_{3}= [^{1}/_{2}] Q_{3}V_{3}= [^{1}/_{2}] ( 1 x 10^{ - 3}C ) (2 V)PE

_{3}= 1 x 10^{ - 2}J = 1.0 mJNotice that

PE _{Tot}= 9 mJ = 7.5 mJ + 0.5 mJ + 1.0 mJ = PE_{1}+ PE_{2}+ PE_{3}and that

Resistors R_{2}and R_{3}are inparallelso we begin by replacing them with an equivalent resistor R_{eq}whose value we can calculate from1/R _{eq}= 1/R_{2}+ 1/R_{3}= 1/(500 ) + 1/(1000 )1/R

_{eq}= (0.003) /R

_{eq}= [1/0.003] = 333.3[ Notice that R

_{eq}< 500 or that R_{eq}< R_{min}; that is, notice that the equivalent resistance R_{eq}isless thanthe smallest (or smaller) of the resistors in parallel. ]Now we have R

_{1}, this R_{eq}, and R_{4}inseriesas in the following diagram,or

The new equivalent resistor Req,2 for this combination of three resistors in series is given by

R _{eq,2}= R_{1}+ R_{eq}+ R_{4}R

_{eq,2}= 100 + 333 + 400R

_{eq,2}= 833The current from the battery is given by

I = V / R _{eq,2}= 12 V / 833 = 0.0144 AThis is the current through R

_{1}and R_{4},I _{1}= I_{4}= 0.0144 AThe power supplied from the battery is

P = I V = (0.0144 A)(12 V) = 0.173 W P

_{bat}= 0.173 WNow we work back and find the power absorbed by each of the resistors (well, at least find the power absorbed by resistor R

_{1}since that was asked for in this question). We know the current through R_{1}and R_{4}so we can find the voltage across them and then find the power absorbed by them.V _{1}= I R_{1}= (0.0144 A)(100 ) = 1.44 VP

_{1}= I V_{1}= (0.0144 A)(1.44 V) = 0.021 WOnly

Iand_{1}( = 0.0144 A)Pwere asked for in this question. We have found those now. The required solution is complete._{1}( = 0.021 W)But we might as well continue and find the current and power for

allof the resistors -- just for "completness".V _{4}= I R_{4}= (0.0144 A)(400 ) = 5.76 VP

_{4}= I V_{4}= (0.0144 A)(5.76V) = 0.0829 WThe voltage across the bank of resistors in parallel -- R

_{2}and R_{3}-- is the 12 V of the battery minus the voltage across R_{1}and R_{4},V _{2}= V_{3}= 12 V - 1.44 V - 5.76 V = 4.80 VP = I V = (V / R) V = V

^{2}/RP

_{2}= (4.80 V)^{2}/ (500 ) = 0.0461 WP

_{3}= (4.80 V)^{2}/ (1000 ) = 0.023 WI

_{2}= V_{2}/R_{2}= 4.80 V / 500 = 0.0096 AI

_{3}= V_{3}/R_{3}= 4.80 V / 1000 = 0.0048 AAs a quick check, see if these absorbed powers equal the battery's output power,

P _{tot}= P_{1}+ P_{2}+ P_{3}+ P_{4}P

_{tot}= (0.021 + 0.0461 + 0.023 + 0.0829 ) WP

_{tot}= 0.173 W = 0.173 W = P_{bat}And that agrees to three significant figures. This agrees with our ideas of energy conservation.

We have calculated all of the currents

Notice, too, that

as we would expect from Kirchoff's Jucntion Rule or simply from charge conservation

a)What is the final charge Q_{f}that will accumulate on the capacitor in this circuit, after switch S is closed, if we wait long enough?R = 2000

C = 500 µF

V = 12 v

b)What is the time constant of this circuit?

c)What charge will there be on the capacitor 2.0 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,V _{f}= 12 vC = Q / V

Q

_{f}= C V_{f}Q

_{f}= ( 500 x 10^{ - 6}F ) ( 12 v )

Q_{f}= 6.0 x 10^{ - 3}CThe time constant of this circuit is

= RC = ( 2000 ) (500 x 10

^{ - 6}f )

= 1.0 sThe charge on the capacitor, as a function of time, is

q = q(t) = Q_{f }[ 1 - e^{- t / }]

q ( 2.0 s) = q (t = 2.0 s) =Q_{f }[ 1 - e^{- 2.0 / }]

q ( 2.0 s) =Q_{f }[ 1 - e^{- 2.0 / 1.0}]

q ( 2.0 s) =Q_{f }[ 1 - ( e^{- 2}) ]

q ( 2.0 s) =Q_{f }[ 1 - ( 0.135 ) ]

q ( 2.0 s) =Q_{f }[ 0.865 ]

q ( 2.0 s) =0.865 Q_{f }

q ( 2.0 s) =0.865( 6.0 x 10^{ - 3}C)