## Hour Exam #1

## September 20, 1999

Possibly useful information:k = 9 x 10^{9}N m^{2}/C^{2}e = 1.6 x 10

^{ - 19}CC

_{eq}= C_{1}+ C_{2}+ C_{3}P = IV = I

^{2}R = V^{2}/RR

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

1.What is the electric field at the origin (x = 0, y = 0) when charges Q_{1}= - 10 µC and Q_{2}= - 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

E, the field due to charge Q_{1}_{1}and find its direction. Then we will calculateE, the field due to charge Q_{2}_{2}and find its direction. Then we will find thevectorsum of these two electric fields for the net field.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 vector sum 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/CNow 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 10

^{6}N/C= 18.4°

That is, the net electric field is

E = 28.5 x 10 ^{6}N/C at the angle = 18.4^{o}as shown in the diagram.

2.What is the electric potential at the origin (x = 0, y = 0) due to charges Q_{1}= - 10 µC located at (x = 0, y = 0.10 m) and Q_{2}= - 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

V

_{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}voltsV

_{tot}= - 3.6 x 10^{6}volts

3.Three capacitors are connected in the circuit shown below. They have values of C_{1}= 100 F, C_{2}= 200 F, and C_{3}= 300 F. They are connected to a 12-volt battery as shown.Find the charge on each capacitor and the potential energy stored in capacitor C

_{1}.First, we replace the

parallelcapacitors, C_{2}and C_{3}, with a single equivalent capacitor which we will call C_{eq1}:C

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

eq1= (200 + 300) FC

_{eq1}= 500 FNow we replace these

seriescapacitors, C_{1}and C_{eq1}, with a single equivalent capacitor which we will call C_{eq}:C

_{eq}= 83.3 FC = Q/V ===> Q = C V

Q = (83.3 x 10

^{ - 6}F)(12 V)Q = 1 x 10

^{ - 3}C

PE

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

_{Tot}= 6 x 10^{ - 3}J = 6 mJ

Now we return to the

individualcapacitors,Q _{1}= Q = 1 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}= 2.0 VPE

_{1}= [^{1}/_{2}] Q_{1}V_{1}= 5 x 10^{ - 3}J = 5 mJV

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

_{2}isnot12.0 V!Q _{2}= C_{2}V_{2}= (200 x 10^{ - 6}F) (2.0 V) = 4 x 10^{ - 4}C = Q_{2}Q

_{2}= 4 x 10^{ - 4}CPE

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

_{2}= 4 x 10^{ - 4}J = 0.4 mJV

_{3}= 2.0 VQ

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

_{3}= (300 x 10^{ - 6}F) (2.0 V) = 6 x 10^{ - 4}CPE

_{3}= [^{1}/_{2}] Q_{3}V_{3}= [^{1}/_{2}] ( 6 x 10^{ - 4}C ) (2 V)PE

_{3}= 6 x 10^{ - 4}J = 0.6 mJNotice that

PE _{Tot}= 6 mJ = 5 mJ + 0.4 mJ + 0.6 mJ = PE_{1}+ PE_{2}+ PE_{3}and that

Q = Q _{1}= 1 x 10^{ - 3}C = 4 x 10^{ - 4}C + 6 x 10^{ - 4}C = Q_{2}+ Q_{3}

4.Find the equivalent resistance in the circuit shown below for R_{1}= 100 , R_{2}= 500 , R_{3}= 1000 , R_{4}= 50 . If these are connected to a 12-volt battery, what is the power supplied to the equivalent resistor? Find the current through resistor R_{2}and the power supplied to R_{2}.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 + 1/10001/R

_{eq}= (0.003) /R

_{eq}= [1/0.003] = 333.3Now 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 + 50R

_{eq,2}= 483The current from the battery is given by

I = V / R _{eq,2}= 12 V / 483 = 0.0248 AThis is the current through R

_{1}and R_{4},I _{1}= I_{4}= 0.0248 AThe power supplied from the battery is

P = I V = (0.0248 A)(12 V) = 0.2981 W = 0.30 W P

_{bat}= 0.30 WNow we work back and find the power absorbed by each of the resistors. 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}= (0.0248 A)(100 ) = 2.48 VP

_{1}= I V_{1}= (0.0248 A)(2.48 V) = 0.0615 WV

_{4}= I R_{4}= (0.0248 A)(50 ) = 1.24 VP

_{4}= I V_{4}= (0.0248 A)(1.24 V) = 0.0307 WThe voltage across the 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 - 2.48 V - 1.24 V = 8.28 VP = I V = (V / R) V = V

^{2}/RP

_{2}= (8.28 V)^{2}/ (500 ) = 0.137 WP

_{3}= (8.28 V)^{2}/ (1000 ) = 0.069 WI

_{2}= V_{2}/R_{2}= 8.28 V / 500 = 0.0166 AI

_{3}= V_{3}/R_{3}= 8.28 V / 1000 = 0.00828 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}= (0.0615 + 0.069 + 0.137 + 0.0307) WP

_{tot}= 0.2982 W = 0.2891 W = P_{bat}And that agrees to three significant figures.

Now calculate the currents through R

_{2}and R_{3},I _{2}= V_{2}/ R_{2}= 8.28 V / 500 = 0.0166 AI

_{3}= V_{3}/ R_{3}= 8.28 V / 1000 = 0.00828 ANotice, too, that

I _{1}= I_{4}= I_{2}+ I_{3}= 0.0248 A

5. 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?

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,

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 = ( 1000 ) (500 x 10

^{ - 6}f )= 0.5 s

The charge on the capacitor, as a function of time, is

q = q(t) = Q _{f }[ 1 - e^{- t / }]q ( 1.0 s) = q (t = 1.0 s) = Q

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

_{f }[ 1 - e^{- 1.0 / 0.5}]q ( 1.0 s) = Q

_{f }[ 1 - ( e^{- 2}) ]q ( 1.0 s) = Q

_{f }[ 1 - ( 0.135 ) ]q ( 1.0 s) = Q

_{f }[ 0.865 ]q ( 1.0 s) = 0.865 Q

_{f }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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