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PHYS 104
Form A Taibah University - Yanbu Branch Faculty of Sciences and Arts at Yanbu Department of Physics
Physics (2): Midterm Exam
Monday 02nd^ May 2011 AD, 28th^ Jumada al-awwal 1432 AH 16:00 a.m to 18:00 a.m Duration of Exam: Two Hours
Answer ALL questions in the answer sheet Electronic calculators may be used, provided that they cannot store text
USE OF MOBILE PHONES IS STRICTLY PROHIBITED, EVEN AS CALCULATORS A list of relevant constants and equations is provided
Each multiple-choice question is marked out of 1 The total mark for this exam is 20
YOU MAY FIND THE FOLLOWING USEFUL
F^ ~ = qQ 4 πǫ 0
d^ ˆ | d~ |^2 F = k qQ d 2 ,
E~ = Q 4 πǫ 0
d^ ˆ | d~ |^2 E = k dQ 2 , F~ = q E,~ p = qd, from − to + ~τ = ~p × E~ = pE sin φ nˆ U = −p. ~E~ = −pE cos φ ΦE =
E.d^ ~ A~ ΦE = q ǫenc 0 Vab = Va − Vb = −
∫ (^) a b
E.d~^ ~ r Vab = Ed (uniform field)
Er = −dV dr C = Q V 1 Cs^ =^
Ci^ ,^ Series Cp =
Ci, Parallel I = dQ dt V = IR Rs =
Ri, Series 1 Rp^ =^
Ri^ ,^ Parallel ∑^ P =^ IV I = 0 in a junction ∑ V = 0 in a loop I = I 0 exp(−t/τ ), charging τ = RC
PHYSICAL CONSTANTS AND CONVERSION FACTORS
SYMBOL DESCRIPTION NUMERICAL VALUE
n nano 10 −^9 μ micro 10 −^6 m milli 10 −^3 e electric charge 1.602× 10 −^19 C ǫ 0 permittivity of free space 8. 854 × 10 −^12 C^2 N−^1 m−^2 k (^4) πǫ^10 8.988 × 109 N.m^2 /C^2
(b) Rp = 0.417Ω (c) Rp = 2.4Ω (d) Rp = 24Ω
- Two capacitors C 1 = 4 F and C 2 = 6 F are attached in parallel. The equivalent capacitance of the network Cp is: (a) Cp = 10 F (b) Cp = 2.4 F (c) Cp = 0.417 F (d) Cp = 24 F
- A cube encloses two charges +1 C and −2 C. What is the total flux of electric field out of the cube? (a) 0 (b) +1. 13 × 1011 N m^2 /C (c) − 1. 13 × 1011 N m^2 /C (d) −1 C
- A conductor carries charge Q at equilibrium. What is the electric field inside the conductor? (a) 1 N/C (b) 0 (c) Q (d) Q/ 4 πǫ 0
- Four electric wires meet at a junction. The first carries 4 A into the junction, the second carries 5 A out of the junction, and the third carries 2 A out of the junction. The fourth carries: (a) 7 A out of the junction (b) 7 A into the junction (c) 3 A out of the junction (d) 3 A into the junction
- A 200-N/C electric field is in the positive x-direction. The electric force on an electron of charge q = − 1. 6 × 10 −^19 C in this field is: (a) 200 N in the positive x-direction (b) 200 N in the negative x-direction (c) 3.2 × 10 −^17 N in the negative x-direction (d) 3.2 × 10 −^17 N in the positive x-direction
- A wire carries a steady current of I = 2 A. The charge that passes through a cross section of the wire in t = 2 s is: (a) Q = 6. 4 × 10 −^19 C (b) Q = 1 C (c) Q = 4 C (d) Q = 2 C
- For the network of resistors shown in figure 1, what is the equivalent resistance?
1Ω
1Ω 1Ω
1Ω a (^) b
Figure 1: Question 13
(a) 4 Ω (b) 2 Ω (c) 0.5 Ω (d) 1 Ω
- In figure 2, the switch S is initially open. The am-metre reads 2.0 A. When the switch S is closed the current in the am-metre becomes:
b b
20Ω 60Ω
S
15Ω
A
70 V
Figure 2: Questions 14 - 16
(a) 0 (b) 1 A
a
b
Figure 4: Questions 18 - 20
(a) 0 (b) (^4) πǫQ 01 r
(c) (^4) πǫQ 0 r^12
(d) (^4) πǫQ 0 r^13
- What is the electric potential difference across the capacitor: Vab = −
∫ (^) a b E(r)dr? You may assume that (^) ∫ 1 r^2 dr^ =^ −
r. (a) 0 (b) (^4) πǫQ 0
a^3 −^
b^3
(c) (^4) πǫQ 0
a^2 −^
b^2
(d) (^4) πǫQ 0
a −^
b
- What is the capacitance of the capacitor C = Q/Vab
(a) 4πǫ 0 ab/(b − a) (b) 4πǫ 0 (b − a)/ab (c) 0 (d) (^4) πǫ^10 b− aba
END OF EXAM
