The Petroleum Institute
PHYS 241 Physics II
Electromagnetism and Optics
Test 3
Spring Semester 2016
Time allowed: 60 minutes
Direction to Students:
All questions are to be attempted.
All answers are to be written in the space provided below each part of e
1 31. The energy stored by a capacitor is given by U = 2 CV 2 , where V is the potential difference across its plates. We convert the given value of the energy to Joules. Since 1 J = 1 W s, we multiply by (103 W/kW)(3600 s/h) to obtain 10 kW h = 3.6 107 J
33. The energy per unit volume is 1 1 e u = 0E 2 = 0 2 2 4 0 r 2
FG H
IJ K
2
=
e2 . 32 2 0r 4
(a) At r = 1.00 103 m , with e = 1.60 1019 C and 0 = 8.85 1012 C2 /N m 2 , we have u = 9.16 1018 J/m3 . (b) Similarly, at r = 1.00 106 m , u = 9.16 106 J/m3 . (c
34. (a) The potential difference across C1 (the same as across C2) is given by V1 = V2 =
(15.0 F ) (100V ) = 50.0V. C3V = C1 + C2 + C3 10.0 F+5.00 F+15.0 F
Also, V3 = V V1 = V V2 = 100 V 50.0 V = 50.0 V. Thus, q1 = C1V1 = (10.0 F ) ( 50.0V ) = 5.00 104 C
35. (a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by 0 A d i , the charge is q = CV = 0 AVi d i . After the plates are pulled apart, their separation is d f and the potential difference is Vf.
13. (a) and (b) The original potential difference V1 across C1 is V1 = CeqV C1 + C2 =
( 3.16 F ) (100.0 V ) = 21.1V.
10.0 F + 5.00 F
Thus V1 = 100.0 V 21.1 V = 78.9 V and q1 = C1V1 = (10.0 F)(78.9 V) = 7.89 104 C.
16. We note that the voltage across C3 is V3 = (12 V 2 V 5 V ) = 5 V. Thus, its charge is q3 = C3 V3 = 4 C. (a) Therefore, since C1, C2 and C3 are in series (so they have the same charge), then 4 C C1 = 2 V = 2.0 F . (b) Similarly, C2 = 4/5 = 0.80 F.
20. We do not employ energy conservation since, in reaching equilibrium, some energy is dissipated either as heat or radio waves. Charge is conserved; therefore, if Q = C1Vbat = 100 C, and q1, q2 and q3 are the charges on C1, C2 and C3 after the switch is
21. Eq. 23-14 applies to each of these capacitors. Bearing in mind that = q/A, we find the total charge to be qtotal = q1 + q2 = 1 A1 + 2 A2 = o E1 A1 + o E2 A2 = 3.6 pC where we have been careful to convert cm2 to m2 by dividing by 104.
5. We use Gauss law: 0 = q , where is the total flux through the cube surface and q is the net charge inside the cube. Thus, 1.8 106 C = = = 2.0 105 N m 2 C. 2 2 12 0 8.85 10 C N m q
2. We use = E dA and note that the side length of the cube is (3.0 m1.0 m) = 2.0 m. (a) On the top face of the cube y = 2.0 m and dA = ( dA ) . Therefore, we have j
2 E = 4i 3 ( 2.0 ) + 2 = 4i 18j . Thus the flux is j
z
(
)
=
top
E dA =
top
( 4i 18j) (
68. We denote the electron with subscript e and the proton with p. From the figure below we see that e Ee = E p = 4 0 d 2 where d = 2.0 106 m. We note that the components along the y axis cancel during the vector summation. With k = 1/40 and = 60 , the ma
41. We combine Eq. 22-9 and Eq. 22-28 (in absolute values).
F= qE= q
FG p IJ = 2kep H 2 z K z
3 3 0
where we have used Eq. 21-5 for the constant k in the last step. Thus, we obtain F= 2 ( 8.99 109 N m 2 C2 ) (1.60 1019 C ) ( 3.6 1029 C m )
( 25 10
9
m)
3