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QUIZ 18, CH. 18: TEST1 IS TBA

QUIZ 18

EVEN ANSWERS

SELECTED DISCUSSIONS TO exercises/problems BELOW.
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. Each sublist is separated by a semi-colon ";" and includes the name of a representative example for those exercises.

SELECTED DISCUSSIONS TO exercises/problems BELOW. I've try to include interesting explanatory hyperlinks throughout. YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. Each sublist is separated by a semi-colon ";" and includes a representative example for exercises. WE WILL REVIEW THEM IN-CLASS THE LAST WEEK BEFORE EXAMS.

CH. 18 MULTIPLE CHOICE 5, 6, 8, 10, 11, 14, 15 - Exercise/problems example 1:1, 2, example 2:3, example 3:22, 23, 24, example 4:17, example 5:28, example 7: 37, 38, example 6: 41, example 8: 49, 54, example 9: 81, 84, example 10: 68, 69

TURN IN: 2, 3, 22, 24, 17, 28, 37, 41, 54, 81, 68.

* DISCUSSIONS PROVIDED.

2. work done by field = = -eE*d

I'd like to clarify something I said in class.

IF YOU WANT YOU CAN solve for final speed:
(1/2)*m*v_f² - (1/2)*m*v_i² = -eE*d.

Solve for v_f .

YOU MAY NOW PROCEED TO BOOK REQUESTED PART OF PROBLEM.

3. Electric potential energy = charge*potential. Let potential V be produced by the -7.70 uC charge; let us call this charge Q<0. Thus potential energy = q*V, where q = 2.30 uC. Note: V = kQ/r. Can you explain why the energy is negative??

22. (a) V_A = kq₁/(0.050 m) + kq₂/(0.050m) (b) V_B = kq₁/(0.080 m) + kq₂/(0.060m)
(c) Work by field = -change in potential energy.

Change in potential energy = charge*(change in potential) = charge*(V_A - V_B ) . Note: charge = 2.50 x10 ^-9 C.
Can you explain the sign of Work by field ?

24. (a) Use conservation of energy: KE_i + U_i = KE_f + U_f, where U represents the potential energy (Book notation-- chapter 7) . KE_i = 0 and U_i= charge*potential. Let potential V be from either charge: V = ke/r, where e is the proton charge: e = -(electron charge) = -(-e) = +e.
U_i = ke²/r. Note the charge squared! U_f = 0 since the charges are infinitely far way from each other.
KE_f = 2*(1/2)mv². Solve for v given m the proton mass.

(b) Force = ke²/r² = m*a. Max a when r is the smallest, the initial separation

17. E = slope of the line. Remember to convert cm to meters.

28. (a) Surface is a plane and it is at the center, between plates, parallel to the two plates.
(b) Surface is a plane and it is (2/12)*6 cm to the right of neg. plate.
(c) E = |gradient| = |total change in V/ total distance|

37. C = Area*epsilon/d (b) V = Q/C (c) Note E*d = V where V is the potential difference between plates.

41. Do you n know the formula for the area of a circle of radius r?
(a) Q = C*V. Note: uC means microfarad, 10^-6 F
(b) Q' = C'V, where V is the same; V = 12 (V).
C' = area*epsilon/(2d) . Get Q' = C'*V
(c) C'' = (new area)*epsilon/d, where new area = pi*(new radius)². Note: new radius = 2*(old radius); get new Q using same method as previous parts.

54. Note:
(1)The two capacitors in series near the a - terminal can be combined to produce an net capacitance = C/2. You should be able to prove this.
(2) Thus, closest to terminal a   we have two capacitors in parallel, C/2 and C. The net capacitance is C/2 + C = (3/2)*C. (3) Thus we finally have two capacitors in series between terminals a and b, C and (3/2)*C. The C_s (s means series) is given by 1/C_s = 1/C   + 1/[(3/2)*C] = 1/C + 2/(3C). Add these up to get the final net-C.
Now, let us find the charges:

Go to step 3 . Before the final combo is computed, you have two C's in series, C and (3/2)*C.   Thus they have the same charge Q.
28 (V) = Q/C   + Q/[(3/2)*C]. Solve for Q, the charge on the bottom C in the figure.
To get the other charges: Note: Voltage across the (3/2)*C is Q/[(3/2)*C], where Q is known. Thus , you can now get the voltage = Q/[(3/2)*C] = 2Q/(3*C). This voltage equals the voltage across the single C in that branch:
2Q/(3*C) = voltage across single C = Q'/C. Find Q' = charge on single C .

FINALLY GET THE CHARGE ON THE ORIGINAL TWO CAPACITORS IN SERIES IN THE UPPER BRANCH:
Since they are in series that have the SAME charge Q". The voltage across these two C's is
2Q/(3*C) = Q"/C + Q"/C.
Find Q".

81.
(a) C = Area*epsilon/d
(b) Q = C*V, where V = V₊   - V_- , the voltage difference between the two plates.
(c) E*d = V
(d) U = Q²/2C.
(e) If the plates are disconnected, you will still have the same charge Q you had in the beginning, computed in part (b).
The force between the plates will be attractive so you would have to pull the plates apart with a force, equal in   magnitude to the attractive one, to keep them from accelerating toward zero separation.
    C' =   C = Area*epsilon/d', where d' is the new separation.   You can immediately find U = Q²/2C'.   Note since C' < C, you have larger energy in the capacitor. That's because you did work to separate the plates, thus increasing the potential energy of the plates. It's like lifting a ball above the ground to increase its U = mgh. You can now find V' = Q/C', where C' is the new capacitance. Note the voltage difference has increased to V'. Finally, find E = V'/d' and you should note the magnitude of the electric field E has not changed, also seen formula E = Q/(A*epsilon).    But you are increasing the energy in the space between the plates because the volume of that region has grown. Remember the formula
U = (1/2)*epsilon*E²*volume , where the energy density is (1/2)*epsilon*E² and the volume is
A*(separation distance). See equation 18.20.

68. We will review this more in class. The main issue is this. The electric field inside the region between plates where a dielectric resides is smaller than it would be in a vacuum. That's because surface-charges of opposite sign and polarity than plate's charges gather on the edges of the dielectric next to plates. These "edge" charges cause a field in the opposite direction that partially cancels out the vacuum field Eo. See figure 18.33 and class notes:
E = Eo/k, where Eo is the vacuum field if there was no dielectric. C = Q/V = Q/[(Eo/k)*d]. Since Eo = Q/(A*epsilon), we see C = k*Co, where Co is the capacitance without a dielectric. C goes up since V = (Eo/k)*d goes down.
(a) E = Eo < 30000 V/m. Use Eo = Q/(A*epsilon) to get Q.
(b) E = Eo/k < 30000 V/m. . Solve for limit on Eo and use Eo = Q'/(A*epsilon) to get limit Q', a larger charge value than Q..