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Midterm Exam 2 Solutions - Fundamentals of Physics II | PHYS 102, Exams of Physics

Material Type: Exam; Class: Fundamentals of Physics II; Subject: Physics; University: Drexel University; Term: Spring 2008;

Typology: Exams

2010/2011

Uploaded on 06/07/2011

justinc97
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PHYS-102 SPRING-08 QUIZ-II Time Limit: 50 min.
______________________________________________________________________________
Name______________________ Recitation Sec.#_______
Notes:1. For problems 2 and 3, your solutions must have adequate details to get full credit.
F = qE
1 2
2
kQ Q
Fr
=
1/4πε0 =
9 2 2
9 10k x Nm C !
=
U = kQ1Q2/r ; V = kQ/r ; 1/C = 1/C1+ 1/C2+1/C3+…. C= Q/V
C = C1 + C2 +C3+…. U = QV/2 ;
Δ
Wfield = -
Δ
U ; Ex = - dV/dx
____________________________________________________________________
NOTE: For Prob .1, below, please circle the answer of your choice for each part.
1a.( 3pts.) Suppose at one instant the electrons and the
nucleus of a helium atom occupy the positions as shown.
The electric potential energy of this charge configuratio n
is:
[a] – 3.5kQ2/r [b] – 4kQ2/r [c] zero [d] + 4kQ2/r
1b. ( 3pts.) Two capacitors are connected in series as shown with
C1=2C2. If U1 and U2 are, respectively the potential energy stored in
C1 and C2., which of the following statements is true?
[a] U1 = 2U2 [b] 2U1 = U2 [c] U1 = U2 [c] 4U1 = U2
1c. ( 3pts.) When the potential difference bet ween the plates of a capacitor is doubled, the
potential energy stored in the capacitor is:
[a] doubled [b] halved [c] unchanged [d] quadrupled
1d. ( 3pts.)Consider two conducting metallic spheres S1 and S2. The radius of S1 is half that of
S2. Each sphere is given some excess charge such that both spheres have the same potential at
their surfaces. This implies that:
[a] S1 and S2 have equal charges.
[b] S2 has more charge on it than S1 .
[c] S1 has more charge on it than S2
r r
-Q +2Q -Q
C1C2
!=25 V
pf3

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PHYS- 102 SPRING- 08 QUIZ-II Time Limit: 50 min.

______________________________________________________________________________ Name______________________ Recitation Sec.#_______ Notes : 1. For problems 2 and 3, your solutions must have adequate details to get full credit. F = q E^12 kQ Q F r = 1/ 4 πε 0 = k = 9 x 1 09 Nm^2^ C!^2 U = kQ 1 Q 2 /r ; V = kQ/r ; 1/C = 1/C 1 + 1/C 2 +1/C 3 +…. C= Q/V C = C 1 + C 2 +C 3 +…. U = QV/2 ; Δ Wfield = - Δ _U ; Ex = - dV/dx


NOTE: For Prob.1, below, please circle the answer of your choice for each part. 1a .( 3pts .) Suppose at one instant the electrons and the nucleus of a helium atom occupy the positions as shown. The electric potential energy of this charge configuration is: [a] – 3 .5kQ^2 /r [b] – 4 kQ^2 /r [c] zero [d] + 4kQ^2 /r 1b. ( 3pts. ) Two capacitors are connected in series as shown with C 1 =2C 2. If U 1 and U 2 are, respectively the potential energy stored in C 1 and C 2 ., which of the following statements is true? [a] U 1 = 2U 2 [b] 2 U 1 = U 2 [c] U 1 = U 2 [c] 4U 1 = U 2 1c. ( 3pts. ) When the potential difference between the plates of a capacitor is doubled, the potential energy stored in the capacitor is: [a] doubled [b] halved [c] unchanged [d] quadrupled 1d. ( 3pts .)Consider two conducting metallic spheres S1 and S2. The radius of S1 is half that of S2. Each sphere is given some excess charge such that both spheres have the same potential at their surfaces. This implies that: [a] S1 and S2 have equal charges. [b] S2 has more charge on it than S. [c] S 1 has more charge on it than S 2 r r

  • Q + 2 Q - Q C 1 C 2 != 25 V

Prob.2. Three charges are arranged on the three corners of a rectangle as shown (Q = 2.0x

  • 6 C) . [a] ( 4 pts ) Determine the electric potential energy of the 3 - charge configuration. 2 3 2 3 2
  1. 1 0. 08 0. 06 kQ kQ kQ U

= kQ^2 ( - 10.0+37.5 – 50) = - 22.5 kQ^2 = - 22.5910^9 4.010-^12 = - 0.81 J [b] ( 3pts ) Determine the electric potential due to the three-charge configuration at P, the unoccupied corner of the rectangle. Potential, V(P) = 3kQ/0.1 – kQ/0.08 + kQ/0.06 = kQ( 30 - 12.5 +16.7) = 6.15*10^5 V [c] ( 3 pts )How much work would you have to do on a fourth charge q = 5.0 nC to bring it from infinity to P? ! W (^) e (^) xt =! U = q! V = q [ V ( P ) " V ( # )] = qV ( P )= 5.010-^9 * 6.15*10^5 = 3.1x10-^3 J

P

- Q

3 Q

Q

6. 0 cm

8. 0 cm 10. 0 cm