Download Quiz Solutions for PHYS-102 Spring-09, QUIZ-II in Physics - Prof. Som D. Tyagi and more Exams Physics in PDF only on Docsity!
OUTLINE OF SOLUTIONS
PHYS-102 SPRING-09 QUIZ-II Time Limit: 50 min.
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Name______________________ ____ Recitation. Sec._______
NOTES:1. For Prob.1, please circle the answer of your choice for each part.
2. For problems 2 and 3, your solutions must have adequate details to get full credit. 3. The electric potential of a charge at infinity is taken to be zero.
F = q E
1 2 2
kQ Q F r
; 1/SH = k 9 10 x^9 Nm C^2 ^2 ,
U = kQ 1 Q 2 /r ; V = kQ/r ; Ex = - dV/dx
1/C ser = 1/C 1 + 1/C 2 +1/C 3 +…. ;
C par. = C 1 + C 2 +C 3 +….;
C = Q/V ; U = QV/2 ;
- ' Wfield = ' U = ' _Wext ;
Problem Score
1
2
3
Total
1a .( 3pts .)Consider two conducting metallic spheres S1 and S2. The radius of S1 is half that of S2. Each sphere is given an excess positive charge Q. If the two spheres now are brought in electrical contact:
[a] charge will flow from S1 to S2.( since V = kQ/r, S1 will be at a higher potential than S2)
[b] charge will flow from S2 to S. [c] no charge flow will take place. [d] the charge on each sphere will double.
1b ( 3pts. ) When the potential difference between the plates of a capacitor is doubled, the magnitude of the electric energy stored in the capacitor:
[a] is doubled [b] is halved [c] remains the same [d] is quadrupled ( U = QV/2= CV^2 /2)
1c (3pts) The electric potential, V(x) in a certain region of space is given by the expression V(x) = 3.0 + 4.0x , where x is measured in meters and V(x) in Volts. The value of the E -field at x = 2.0 m will be (in units of V/m ): [a] 11.0 [b] – 5.0 [c] – 4.0 (E x = - dV/dx) [d] 8.
1d(3pts) Two charges are placed along the x- axis as shown. Now, suppose you move the positive charge ( while keeping the negative charge at the origin) to a point 2X away from the origin. As a result of this displacement the electric potential energy of the two charges would: [a] increase [b] decrease [c] remain the same [d] be reduced to zero
(You will have to do work to move the charges and hence increase the potential energy of the two charges. Or you can simply find ' U =Uf - Ui = - kQ^2 /2X+kQ^2 /X = kQ^2 /2X to see that the potential energy increases when you double the separation .)
- Q + Q
X
Prob. 3( 10 pts). Consider the three capacitors connected as shown. A potential difference of Vac = 120V is maintained between points a and c.
[a]( 2 pts) What is the equivalent capacitance between points a and c?
C (^) ab = 4+6 = 10 PF (parallel combination)
1/C (^) ac = 1/Cab + 1/C (^) bc = (1/10 +1/2) PF-1^ (series combination)
C (^) ac = [10*2]/[10+2] = 5/3 = 1.67PF
[b] ( 2 pts) What is the value of the ratio Vbc / Vab? (Here Vbc and Vab are, respectively, the values of the potential difference betweenb andc, and betweena andb.)
Vbc / Vab = Cab /Cbc = 10/2 = 5 …[1]
[c] ( 2 pts) What is the potential difference Vbc across the 2.0 P F capacitor?
Vbc + Vab = 120V …[2]
From eq. [1], Vbc = 5 Vab , substitute this in [2] to get6Vab = 120
orVab = 20 V andVbc = 100 V
[d] ( 2 pts) How much electric energy is stored in the 6.0 P F capacitor?
2 6 2 6 6
U C V^ ab mJ u u
[e] (2 pts)What is the charge on the4.0 PF capacitor?
Q 4 = C 4 Vab = 4.0*10-6^ *20.0 = 80.0PC
a b c
4.0PF
6.0PF
2.0PF
