Exam 1 Solutions - Department of Physics

[Pages:6]PHY2054 Fall 2014

Exam 1 Solutions

Prof. Paul Avery Prof. Andrey Korytov Sep. 26, 2014

1. Charges are arranged on an equilateral triangle of side 5 cm as shown in the

q1

diagram. Given that q1 = 5 ?C and q2 = q3 = -2 ?C find the magnitude of the

net force on charge q1 (in N).

Answer: 62

q2

q3

Solution: The x component of the force cancels and the y components of the

force on q1 are the same for both charges. Let a be the length of each side. Then the total force in

the

y

direction

is

2

q1q2 a2

k

sin 60

=

62

N

.

2. Charges are arranged on a square of side d as shown in the diagram. In what direction does the electric field at the center of the square point? (The quadrants are numbered counterclockwise starting from the positive x-axis.)

Answer: Fourth quadrant

Solution: The electric fields at the origin due to the left & right (+Q and ?Q) and top & bottom (+Q and ?2Q) charges do not cancel. The electric fields due to these two charge pairs are then in the directions +x (for the left & right charges) and ?y (for the top & bottom charges), leading to a net electric field in the 4th quadrant.

3. Two charged particles are fixed to the x-axis: particle 1 of charge q1 = 20 ?C at x = 0 m, and particle 2 of charge q2 = -80 ?C at x = 0.6 m. At what coordinate along the x-axis is the net electric field produced by the particles equal to zero?

Answer: -0.6 m

Solution: Between x = 0 and x = 0.6 m, electric fields due to charges q1 and q2 point in the same direction and cannot cancel. For x > 0, the two fields are in opposite directions, but the larger in magnitude charge q2 is closer and hence its field is always greater than the field due to charge q1; hence, the two fields cannot cancel each other here either. For x < 0, the fields are again in the opposite directions and would cancel, if their magnitudes are the same:

( ) q1

x2

k

=

q2 2 k . L-x

Solving yields x = -L = -0.6 m.

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PHY2054 Fall 2014

4. An electron with velocity v0 along the +x direction enters a region with a uniform electric field of magnitude 5000 V/m in the -y direction. If the electron travels 5.0 cm in the x direction and gets deflected by 1.0 cm in the y direction what is v0 (in m/s)?

Answer: 1.05 ? 107

Solution: The deflection by a constant acceleration is

y=

1 2

at

2

,

where

y

=

0.05

m,

the

acceleration a = Ee / me and the time t = x / v0 . Solving yields v0 = 1.05 ? 107 m/s.

5. A conducting sphere with charge +10 nC is placed at the center of two concentric conducting spherical shells of radius r1 = 2 cm and r2 = 5 cm (measured to the outer surfaces). The inner shell carries a charge of -7 nC and the outer shell has a charge of +6 nC. Find the charge (in nC) on the inner surface of the outer shell.

Answer: -3

Solution: The E field inside the conductor of the outer shell must be 0. A Gaussian surface drawn within the outer shell will thus have no flux going through it, implying, from Gauss' law, that there is 0 charge enclosed by the Gaussian surface. Since the total charge of the central sphere and inner shell is +3 nC, the charge on the inner surface of the outer shell must therefore be -3 nC.

6. In the previous problem, find the magnitude of the electric field at r = 50 cm.

Answer: 324 V/m

Solution: Outside the uniformly charged shells/spheres, the electric field is kq / r2 and points radially. In this problem, the total field due to the two shells and the central sphere is then

E

=

kq1

/

r2

+

kq2

/

r2

+

kq3

/

r2

=

q1

+

q2 r2

+

q3

k

=

Q r2

k

,

where Q = +9 nC. This yields E = 324 V/m/

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PHY2054 Fall 2014

7. Four charges of magnitude Q = 3.0 ?C (but different signs, as in the figure) +Q

-Q

are arranged on the corners of a square of side 25 cm. Find the potential energy

of the system of the four charges (in J).

Answer: -0.84

Solution: For a system of point charges, the total potential energy is

U =

kqiq j

all (i,j) pairs r

-Q

+Q

With four charges, there are 6 pairs: 12, 13, 14, 23, 24, 34. Let L be the length of a side. Four pairs (corresponding to the 4 sides) each contribute -kQ2 / L while the diagonals each contribute kQ2 / 2L . The total potential energy is thus

( )

4

-

kQ2 L

+

2

kQ2 2L

=

kQ2 L

-4 +

2 = -0.84 J .

8. Electrons in a particle beam have a kinetic energy of 3.2 ? 10-17 J. What is the magnitude of the electric field (in V/m) that will stop these electrons in a distance of 0.1 m?

Answer: 2000 Solution: The distance d can be calculated from K = Fd = Eed , yielding E = 2000 V/m.

9. The movement of a charge in an electric field from one point to another at constant speed without the expenditure of work by or against the field

Answer: none of these

Solution: The work is zero only when moving along an equipotential surface.

10. Two particle each with charge Q are fixed at the vertices of an equilateral triangle with sides of length a. The work required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges is

Answer: 2kQq / a

Solution: The work required of an external force to move the charge q equals to the change in the potential energy of that charge:

Wext

=

U

=U

f

-Ui

=

2!# "

Qq (0.5a)

k

$ & %

-

2!# "

Qq a

k

$ & %

=

2 Qq k a

.

a

a

a

11. Two isolated conducting spheres are separated by a large distance. Sphere 1 has a radius of R and an initial charge 3Q while sphere 2 has a radius of 3R and an initial charge 7Q. A very thin

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PHY2054 Fall 2014

copper wire is now connected to the spheres to allow charge to flow between the spheres. How much charge will be transferred from sphere 2 to sphere 1? (Note that the charge transferred can be positive, negative or zero.)

Answer: -Q / 2

Solution: After reaching equilibrium, the two spheres must be at the same potential, thus

Q1 R

k

=

Q2 3R

k

Let Q be the charge taken from sphere 2 and moved to sphere 1. Then:

3Q

+ Q R

k

=

7Q - Q 3R

k

from where we obtain Q = -Q / 2 .

12. Four protons are placed at rest on the vertices of a square of side 2.0 ?m. The protons are released simultaneously. When the protons are very far away, what is their speed (in m/s)?

Answer: 430

Solution: Conservation of total energy yields 0 + Ui = K f + 0 , where the initial potential energy

( )( ) Ui = ke2 / L 4 + 2 (see problem 7) and the final kinetic energy of four protons

( ) ( )( ) K = 4

1 2

mv

2

. From here, we obtain

v=

ke2 / mL 2 + 1/ 2 = 432 m/s .

13. Two equipotential surfaces lying near the middle of the space between the plates of a parallel-plate capacitor are 2.0 mm apart and have a potential difference of 0.0012 volt. The area of each plate is 7.5 cm2. What is the magnitude of the charge on each plate, (in units of 10-15 C)?

Answer: 4

Solution: The electric field between two uniformly charged plates with charges Q and -Q is E = / 0 = Q / 0 A , where A is the area of each plate. We calculate E = V / d = 0.6 V/m. Solving for Q yields Q = 4.0 ? 10-15 C .

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PHY2054 Fall 2014

14. A certain parallel plate capacitor with capacitance 12 ?F is connected to a source of EMF with potential 3 V. A dielectric material with = 4 is then inserted between the plates of the capacitor with the capacitor still connected to the circuit. By how much does the energy stored in the capacitor change?

Answer: 1.6 ? 10-4 J

Solution:

The

energy

in

a

capacitor

is

U

=

1 2

CV

2

,

with

V

the

voltage

across

it.

Since

the

capacitor remains connected to the emf source when the dielectric is inserted, the change in

( ) energy is caused only by the change in capacitance C = C f - Ci = C - C = -1 C . Thus the

( ) change in energy is U = 0.5 ? 3? 12 ? 10-6 ? 32 = 162 ?J .

15. An air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 4 pF. The dielectric constant of the wax is:

Answer: 4.0

Solution: The capacitance of a parallel-plate capacitor is

C

= 0

A d

Without the dielectric, the doubling of the plate spacing would change the capacitance from 2 pF to 1 pF. Thus to achieve 4 pF, the dielectric constant must be 4.0.

16. A parallel plate capacitor with a capacitance of 2.0 nF is charged to have 0.8 ?C on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?

Answer: +1.6 ? 10-4 J

Solution: If the plate separation is doubled, Cnew = C0 / 2 . With the charge unchanged, the change in energy is thus

U = Q2 - Q2 = Q2 = +160 ?J . 2Cnew 2C0 2C0

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PHY2054 Fall 2014

17. A 9 V battery is connected to a 3 resistor. How much charge passes through the resistor in 3 hours? Answer: 32,400 C Solution: The total charge Q = i ? t , where i = 3 A and t = 10,800 s. Thus Q = 32,400 C.

18. You have two resistors R1 and R2, made from the same material, with the length and diameter of R1 both double that of R2. If R1 and R2 are connected in parallel to a battery, find the ratio of the currents running through R1 and R2 (find I1 / I2).

Answer: 2

Solution:

The

resistance of

a

cylindrical

resistor

is

R=

L A

=

L (d / 2)2

.

The

cross sectional area

of R1 is 4 times that of R2 while its length is double, therefore

R1 =

1 2

R2

.

The

current

in

R1

is

thus

twice the current in R2.

19. In the circuit diagram shown find the voltage drop across the 3.0

resistor.

12 V

Answer: 4.5 V

6.0 4.0

12

Solution: There is 12 V applied across the bottom branch, whose total

3.0

5.0

resistance is 3 + 5 = 8 (resistors connected in series). Thus the

current is 12 V / 8 = 1.5 A and the voltage drop across the 3 resistor is 1.5 A ? 3 = 4.5 V.

20. In the circuit shown find the total energy (in ?J) stored on all the

capacitors. Where C1 = 2 ?F, C2 = 4 ?F, C3 = 9 ?F, C4 = 1 ?F, C5 = 3

?F, and V = 10V.

V

Answer: 182

C2

C1

C4

C5

C3

Solution: The total energy can be calculated using the equivalent capacitance and the emf since

from the battery's perspective the equivalent capacitance behaves in exactly the same way as the

original capacitors. C4 and C5 combine to give C45 = 3.0 ?F. Then C2, C45, C3 in series give 1.64 ?F. Adding in C1 in parallel then gives a total capacitance of 3.64 ?F. The energy is then

U

=

1 2

CV

2

=

1 2

?

3.64

? 102

=

182 ?J

.

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