Q1. A metal plate is connected by a conducting wire to the ground ...

Phys102 Second Major-153 Zero Version

Coordinator: Dr. A. Naqvi Thursday, August 18, 2016 Page: 1



Q1.

A metal plate is connected by a conducting wire to the ground through a switch as

shown in FIGURE 1. The switch is initially closed. A charge +Q is brought close to

the plate without touching it, and then the switch is opened. What is the charge on the

plate after the switch is opened and the charge +Q is removed?

A) −Q

B) +Q

C) 0

D) −Q/2

E) −2Q

Ans:

A

Q2.

Two small identical metal spheres with negligible mass carry charges q1 and q2 with

opposite sign of charges such that │q1│> │q2│. When the charges are separated by a

distance of 1.0 m, they experience a 2.5 N attractive force. Both charges are then

brought in physical contact till the electrostatic equilibrium is reached among the two

spheres. Finally they are separated from each other and placed at a distance of 1.0 m

apart. Now they experience a 2.5 N repulsive force. What were the magnitudes of the

original charge q1 and q2 respectively?

A) 40.2 C, 6.90 C

B) 50.2 C, 13.7 C

C) 4.51 C, 2.85 C

D) 15.5 C, 9.67 C

E) 59.7 C, 10.5 C

Ans:

F1 = kq1q2

d2; F2 =

k(q1 − q2)2

d2; q1q2 =

F1𝑑2

𝑘; q1 − q2 = √

F2𝑑2

𝑘

q1q2 = 2.777 × 10−10C2; q1 − q2 = 3.333 × 10−5 C; q1 − q2 + 3.333 × 10−5 C

q1q2 = q2 (q2 + 3.333 × 10−5) = 2.777 × 10−10 C2

𝑞22 + 3.333 × 10−5q2 − 2.777 × 10−10 = 0; q2 = 0.690 × 10−5 𝐶; 4.02 × 10−5 𝐶

q1 = 0.691 × 10−5 + 3.333 × 10−5𝐶 = 4.0233 × 10−5 𝐶

Figure 1



King Fahd University of Petroleum and Minerals

Physics Department c-20-n-20-s-0-e-1-fg-1-fo-0

Q3.

Three identical 1.0 nC positive point charges are placed as shown in FIGURE 2. Find

the magnitude of the resultant electric field due to these three charges at a point P

located 3.0 cm in front of the middle charge.

A) 27 kN/C

B) 15 kN/C

C) 11 kN/C

D) 37 kN/C

E) 45 kN/C

Ans:

θ = tan−1 (1

3) = 18.44° ; Ep = Ex + 2E′cosθ = kq [

1

(0.03)2+

2cosθ

(0.0316)2]

Ep = 9 × 109 × 10−9[1111.1 + 1900.5] = 27104 𝑉/𝑚 = 27 𝑘𝑉/𝑚

Q4.

A proton is shot in a region with a speed 2.0 ×106 m/s making an angle of 30o with +

x-axis. If an electric field E = −112 i kN/C exists in the region, then how far does

the proton travel along the x-axis before reaching its turning point?

A) 14 cm

B) 11 cm

C) 5.1 cm

D) 25 cm

E) 22 cm

Ans:

|𝑎| =𝑞𝑒𝐸

𝑚𝑝=

1.6 × 10−19 × 112 × 103

1.67 × 10−27= 1.07 × 1013𝑚/𝑠

v0𝑥 = v0𝑐𝑜𝑠30 = 2 × 106 × cos 30

𝑇ℎ𝑒𝑛 v 𝑓𝑥2 = 𝑣𝑜𝑥

2 + 2|𝑎|𝑥 ; v𝑓𝑥 = 0

𝑇ℎ𝑒𝑛 𝑥 =v𝑜𝑥

2

2|𝑎|=

(2 × 106 × 𝑐𝑜𝑠30)2

2 × 1.07 × 1013=

3 × 102

2 × 1.07 × 1013= 0.14 𝑚

1.0 cm

1.0 cm

3.0 cm

1.0 nC

1.0 nC

1.0 nC

P

X-axis

Y-axis

𝐸′

𝜃

𝐸𝑥

𝐸′

𝜋

10

Figure 2





Q5.

Rank the potential energies U of four electric dipoles in a uniform electric field E

shown in FIGURE 3 from most positive to most negative.

A) a, c and d tie, b

B) b, a and d tie, c

C) c, b, a and d tie

D) d, a, c and b tie

E) a, c and b tie, d

Ans:

𝑈 = −p ∙ E = −pEcosθ

Ua = −pEcos135° = 0.71 pE

Ub = −pEcos0° = − pE

Uc = −pEcos90° = 0

Ud = −pEcos270° = 0

a, c and d tie, b

Q6.

A point charge located at the origin generates an electric field of magnitude 1.25 × 106

N/C at a point located at a distance of 0.150 m from the origin. What is the electric flux

through the surface of a sphere of 0.150 m radius centered at the origin?

A) 3.53 × 105 N.m2/C

B) 1.22 × 105 N.m2/C

C) 0.97 × 105 N.m2/C

D) 4.55 × 105 N.m2/C

E) 5.11 × 105 N.m2/C

Ans:

ϕ = EA = 1.25 × 106 × 4π(0.15)2 = 3.53 × 105 N ∙ m2/C

a b c d

E

E

E

Figure 3





Q7.

An infinitely wide uncharged conducting metal plate is located in horizontal plane

parallel to an infinitely wide insulating sheet of charge, as shown in FIGURE 4. The

sheet and the metal plate are a distance d apart and the sheet has positive surface charge

density +. What is induced surface charge density on the lower side of the conductor

facing the sheet?

A) /2

B) +/2

C)

D) +

E) 2

Ans:

For lower surface E = σ′

εo

E is due to the sheet , E =σ

2εo

Then E = σ′

εo=

σ

2εo

σ′ = σ

2

Q8.

An infinitely long line of negative charge has a linear charge density C/m and lies

parallel to the y-axis at x = −2.00 m. A positive point charge of magnitude 1.50 C is

located at x =1.00 m and y = 2.00 m. If the magnitude of the x-component of the

resultant electric field at x = 2.00 m and y = 2.00 m is 4.95 × 103 V/m, what is magnitude

of the charge density ?

A) 1.90 × 10-6 C/m

B) 2.31 × 10-6 C/m

C) 0.39 × 10-6 C/m

D) 5.33 × 10-5 C/m

E) 1.00 × 10-5 C/m

Ans:

E𝑝 = 𝐸𝜆 + 𝐸𝑞 = 2𝑘𝜆

4+

𝑘𝑞

(1)2= 4.95 × 103

𝑘𝜆

2+ 𝑘𝑞 = 4.95 × 10−3: 𝜆 = 2 [

4.95 × 10−3

𝑘− 𝑞]

𝜆 = 2

𝑘[4.95 × 103 − 𝑘𝑞] = 2 [

4.95 × 103

9 × 109− 1.5 × 10−6] = −1.90 × 10−6 𝑐/𝑚

Conductor

Conductor lower side

Sheet Sheet surface charge

density +σ

d

𝜎′

Figure 4

E E

q p

C/m

-2m

2m

1 m

(2, -2)

X

Y





Q9.

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere

with radius 0.355 m is 1750 N/C. Assuming the sphere’s charge is uniformly

distributed, what is the electric field inside the sphere at a distance of 0.200 m from

the center?

A) 1.96 × 103 N/C

B) 1.10 × 103 N/C

C) 0.55 × 103 N/C

D) 2.45 × 103 N/C

E) 3.33 × 103 N/C

Ans:

For insulating sphere Er(r > R) = kQ

r2;

Er(r < R) = kQ

R3∙ 𝑟

𝐹𝑜𝑟 𝑟 > 𝑅, 𝐸 (𝑟 > 𝑅) = 1750 =kQ

(0.355 + 0.145)2=

kQ

(0.5)2

⇒ kQ = (0.5)2 × 1750 = 437.5

𝐹𝑜𝑟 𝑟 < 𝑅, 𝐸 (𝑟 < 𝑅) = 𝑘𝑄

𝑅3∙ 𝑟

E′(𝑟 < 𝑅) = 437.5

(0.355)3× 0.2 = 1955.8 𝑁/𝐶 = 1.96 × 103 𝑁/𝐶

Q10.

A proton is placed midway between points A and B. The potential at point A is –20

V, and the potential at point B is + 20 V. The potential at the mid-point is 0 V. The

proton will

A) accelerate toward point A

B) move toward point B with constant velocity

C) accelerate toward point B

D) move toward point A with constant velocity

E) remain at rest

Ans:

B

+20

A

-20

+

E

V= 0





Q11.

An insulating sheet in the XZ plane is uniformly charged with a positive surface charge

density = 3.50 × 10-6C/m2, as shown in FIGURE 5. What is the change in the

potential energy of a point charge Q = 13.6 C when it is moved from point A to point

B.

A) 2.69 J

B) 1.15 J

C) 1.78 J

D) 4.11 J

E) 5.22 J

Ans:

∆y = 2 − 3 = −1 m

∆V = −𝐸𝑦∆y = −𝜎

2𝜀𝑜 × (−1) =

−3.5 × 10−6 × (−1)

2 × 8.85 × 10−12= 197740.1

∆U = q∆V = 13.6 × 10−6 × 197740.1 = 2.689 J

Q12.

Four 1.0-g spheres, each carrying a charge of 5.0 nC, as shown in FIGURE 6, are held

fixed in their positions. They are then released at the same time and allowed to move

away from each other. What is the speed of each sphere when they are very far apart?

Assume the electric potential V = 0 at infinity.

A) 0.11 m/s

B) 1.9 m/s

C) 0.14 m/s

D) 0.02 m/s

E) 2.2 m/s

Ans:

∆U = 4(1

2𝑚v2) = 2 𝑚v2

∆U = 4𝑘𝑞2

0.05+

2𝑘𝑞2

0.0707= 2𝑘𝑞2 [

2

0.05+

1

0.0707]

= 2 × 9 × 109 × (5 × 10−9)2[40 + 14.14] = 2.44 × 10−5

v = √∆𝑈

2𝑚 = √

2.44 × 10−5

2 × 10−3 = 0.11 𝑚/𝑠

Figure 6

Ey 𝐸𝑦

𝐸𝑦 𝑦

Figure 5

50

7.05





Q13.

FIGURE 7 shows two uniformly charged spheres having charges of 100 nC and 25 nC

and separated by a center-to-center distance of 100 cm. What is the potential difference

between points a and b, i.e. Vba? Note that points a and b are located on the surfaces of

the larger and smaller spheres, respectively. Assume V = 0 at infinity.

A) 2.1 kV

B) 1.0 kV

C) 1.5 kV

D) 3.2 kV

E) 4.4 kV

Ans:

Vba = Vb − Va ;

Vb = k [qa

0.95+

qb

0.05] = 9 × 109 [

100

0.95+

25

0.05] × 10−9 = 5447.34 V

Va = k [qa

0.3+

qb

0.7] = 9 × 109 [

100

0.3+

25

0.7] × 10−9 = 332.14 V

Vba = Vb − Va = 5447.3 − 3321.4 = 2125.9 V = 2.13 × 103 V

100 nC

60 cm

100 cm

10 cm

25 nC a b

Figure 7





Q14.

Six identical capacitors with capacitance C = 5 F are connected to a power supply

V= 12 V as shown in FIGURE 8. What is the equivalent capacitance of these six

capacitors and what is the potential difference Vab between points a and b ?

A) 7.5 F ; 0.0 V

B) 1.5 F ; 9.0 V

C) 9.7 F ; 6.5 V

D) 3.3 F ; 2.0 V

E) 12 F ; 0.0 V

Ans:

Ceq = 3 C

2= 3 ×

5

2= 7.5 μF

Vb = Va Vb − Va = 0

Figure 8

⇒ ⇒ V

C C C

C C C

b a

C/2 C/2 V C/2 V 𝐶𝑒𝑞

⇒





Q15.

In FIGURE 9, V = 12 V, C1 = C5 = C6 =6.0 F, and C2 = C3 = C4 = 4.0 F. What is the

charge stored on the capacitor C1 and C5?

A) 48 C , 0 C

B) 16 C, 32 C

C) 22 C, 12 C

D) 65 C, 0 C

E) 16 C, 10 C

Ans:

Ceq = C1 × C234

C1 + C234 where C234 = C2 + C3 + C4 = (4 + 4 + 4)μF = 12μF

Ceq = 6 × 12

6 + 12 μF = 4μF

Q1 = Ceq × V = 4 μF × 12 = 48 μC

Q5 = 0 μC (short circuit)

Figure 9

𝑪𝟐 𝑪𝟑 𝑪𝟒

𝑪𝟏

𝑽

V 𝐶𝑒𝑞

⟹ ⟹

Short Circuit





Q16.

An air capacitor made of two flat square plates, each with area A = 100 cm2 and

separated by a distance d = 10.0 cm, is connected across a 150 V battery. Then a metal

square slab having thickness a = 5.00 cm and area A = 100 cm2 is inserted in the

middle of the air gap between the plates without touching either plate, as shown in

FIGURE 10. What is the energy stored in the circuit now?

A) 20.0 nJ

B) 25.0 nJ

C) 33.3 nJ

D) 15.7 nJ

E) 10.1 nJ

Ans:

C = ε0

A

(d − a

2 )=

8.85 × 10−12 × 100 × 10−4

(10 − 5

2 ) × 10−12= 3.54 × 10−12F

Ceq = C

2=

3.54 × 10−12

2= 1.77 × 10−12μF

𝑈 =1

2 𝐶𝑒𝑞𝑉

2 = 1

2× 1.77 × 10−12 × (150)2 = 1.99 × 10−8 J = 1.99 × 10−9 J

a d

Figure 10

C

C

150 V 𝐶𝑒𝑞 150 V ⇒ ⇒





Q17.

Two parallel plate capacitors C1 and C2 are connected in series to a 96.0 V battery. Both

capacitors have plates with an area of 1.00 cm2 each and a separation of 0.100 mm. C1

has air between the plates and C2 has that space filled with porcelain with a dielectric

constant =7. If the capacitors are connected across the battery for a long time, what is

the charge on capacitor C1?

A) 0.74 nC

B) 1.52 nC

C) 0.52 nC

D) 0.24 nC

E) 2.55 nC

Ans:

Equivalent capacitance C1 = ε0

A

d=

8.85 × 10−12 × 10−4

0.1 × 10−3= 8.85 × 10−12 F

C2 = κ ∙ C1

Ceq = C1 × κC1

C1 + κC1=

κ ∙ C1

κ + 1=

7

8 C1 =

7

8 × 8.85 × 10−12 F = 7.74 × 10−12 F

𝑄𝑐1 = Ceq × V = 7.74 × 10−12 F × 96 = 0.74 × 10−9 C

Q18.

At what temperature will tungsten have a resistivity four times that of copper when

copper is at room temperature (20C)? ( tungsten-room-temp = 5.6 ×10-8 .m; tungsten = 4.5

× 10-3( C)-1 Cu-room-temp = 1.7 x 10-8 .m)

A) 67.6 C

B) 89.1 C

C) 33.2 C

D) 12.2 C

E) 99.5 C

Ans:

ρTung(T) = ρcu−RT (1 + αTungsten∆T) ∆T = (ρTung(T)

ρCu−RT− 1) ×

1

αTung

∆T = (4 × 1.7 × 10−8

5.6 × 10−8− 1) ×

1

4.5 × 10−3= 47.55 C

𝑇𝑓 = 𝑇𝑖 + ∆T = 20 + 47.55 = 67.55℃





Q19.

A copper wire joins an aluminum wire whose diameter is twice that of the copper wire.

The same current flows in both wires. The density of conduction electrons in copper

nCu=1.1×1029 m-3 and the density of conduction electrons in aluminum nAl=2.1 × 1029

m-3. For the two wires the drift speed ratio vd-Cu / vd-Al and the current density ratio

JCu/JAl respectively, are:

A) 7.6, 4.0

B) 3.7, 2.4

C) 1.5, 2.6

D) 11, 5.5

E) 9.6, 2.2

Ans:

nAqvd = I ; vd = I

nAq ;

vd−cu

vd−Al=

nAl × AAl

nCu × ACu

(IAl = ICu)

vd−Cu

vd−Al=

2.1 × 1029 × 4

1.1 × 1029 × 1

JCu

JAl=

IAl

ACu×

AAl

IAl=

AAl

ACu= 4

Q20.

A certain brand of hot-dog cooker applies a potential difference of 120 V to the opposite

ends of the hot-dog and cooks it by means of heat produced in the hot-dog. If 48.0 kJ

of heat is needed to cook each hot-dog, what current is needed to be supplied by the

120 V power supply to cook three hot-dogs simultaneously in 2 min? Assume the hot-

dogs form three resistors connected in parallel in the circuit.

A) 10.0 A

B) 8.22 A

C) 12.1 A

D) 5.72 A

E) 15.1 A

Ans:

𝐸𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑐𝑜𝑜𝑘 3 ℎ𝑜𝑡 − 𝑑𝑜𝑔𝑠 = 3 × 48 × 103 J = 1.44 × 105 J

Time required to cook = 2 minutes

𝑃𝑜𝑤𝑒𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 (J

𝑠) =

1.44 × 105

120= 1200 𝑊

𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 𝑏𝑎𝑡𝑡𝑒𝑟𝑦 = 𝑃

𝑉=

1200

120= 10 𝐴

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