Chapter 22 Solutions - Department of Physics

PHY2049

R. D. Field

Chapter 22 Solutions

Problem 1:

A +15 microC charge is located 40 cm from a +3.0 microC charge. The

magnitude of the electrostatic force on the larger charge and on the smaller

charge (in N) is, respectively,

Answer:

2.5, 2.5

Solution: The magnitiude of the electrostatic for is given by,

KQ1Q2 (8.99 ¡Á 109 Nm 2 / C 2 )(15?C )(3?C )

=

= 2.53N .

F=

r2

(40cm) 2

Remember the force on 1 due to 2 is equal and opposite to the force on 2

due to 1.

Problem 2:

Two point particles have charges q1 and q2 and are separated by a distance

d. Particle q2 experiences an electrostatic force of 12 milliN due to particle

q1. If the charges of both particles are doubled and if the distance between

them is doubled, what is the magnitude of the electrostatic force between

them (in milliN)?

Answer:

12

Solution: The magnitude of the initial electristatic force is

Fi =

Kq1q2

= 12mN .

d2

The magnitude of the final electrostatic force is

Ff =

K (2q1 )(2q2 )

= Fi = 12mN .

( 2d ) 2

+Q

Problem 3:

Two identical point charges +Q are

located on the y-axis at y=+d/2 and

y=-d/2, as shown in the Figure. A third

charge q is placed on the x-axis. At

what distance from the origin is the net

force on q a maximum?

Solutions

Chapter 22

F2

r

q

d

¦È

x

F1

+Q

Page 1 of 6

PHY2049

R. D. Field

d /( 2 2 )

Answer:

Solution: The net force on q is the superposition of the forces from each of

the two charges as follows:

r

KQq

(cos ¦È x? ? sin ¦È y? )

F1 =

2

r

r

KQq

(cos ¦È x? + sin ¦È y? )

F2 =

2

r

r

r

r

2 KQq

cos ¦È x?

F = F1 + F 2 =

2

r

2 KQqx

=

x?

( x 2 + (d / 2) 2 )3 / 2

where I used r2 = x2 + (d/2)2 and cos¦È

¦È = x/r. To find the maximum we

take the derivative of F and set it equal to zero as follows:

?

?

1

3x 2

dF

?

= 2 KQq?? 2

?

2 3/ 2

2

2 5/ 2 ?

+

+

(

(

/

2

)

)

(

(

/

2

)

)

dx

x

d

x

d

?

?

(

)

2 KQq (d / 2) 2 ? 2 x 2

=

=0

( x 2 + (d / 2) 2 )5 / 2

Solving for x gives x = d /(2 2 ) .

Problem 4:

Two 2.0 gram balls hang from lightweight

insulating threads 50 cm long from a

common support point as shown in the

Figure. When equal charges Q are place on

each ball they are repelled, each making an

angle of 10 degrees with the vertical. What

is the magnitude of Q, in microC?

Answer:

0.11

Solution: Requiring the x and y

components of the force to vanish yields

Solutions

Chapter 22

¦È

L

QE Q

T

Q

r

mg

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PHY2049

R. D. Field

KQ2

T sin ¦È = QE = 2

r .

T cos¦È = mg

Eliminating the tension T and solving for Q2 gives

mgr 2 tan ¦È 4 mgL2 sin 2 ¦È tan ¦È

=

Q =

K

K

2

and

Q=

4 ( 2 ¡Á 10 ?3 kg )( 9.8m / s 2 )( 0.5m) 2 sin 2 10 0 tan 10 0

= 0.11?C .

8.99 ¡Á 10 9 Nm 2 / C 2

Problem 5:

Q

-2Q

Four point charges q, Q, -2Q, and Q form a

square with sides of length L as shown in the

Figure. What is the magnitude of the resulting

electrostatic force on the charge q due to the

other three charges?

Answer:

0.41KqQ/L2

L

3

1

Q

L

q

tot

Solution: The electrostatic force on q is the

vector superposition of the following three

forces,

2

r KqQ

F1 = 2 x?

L

r

KqQ

F2 = ? 2 y?

L

r

2 KqQ ? 1

1 ?

?

?

+

F3 =

x

y? ?

?

( 2 L) 2 ?

2

2 ?

and the net force is thus,

r

KqQ ?

1 ?

Ftot = 2 ?1 ?

?( x? ? y? ) .

L ?

2?

The magnitude is given by

r

KqQ

Ftot = 2

L

Solutions

(

)

2 ? 1 = 0.41

Chapter 22

KqQ

.

L2

Page 3 of 6

PHY2049

R. D. Field

Problem 6:

¦È

L=1m

L=1 m

Two 2.0 gram charged balls hang from

lightweight insulating threads 1 m long from a

¦È

T

common support point as shown in the Figure.

Q2

If one of the balls has a charge of 0.01 microC Q1

x = 15 cm

and if the balls are separated by a distance of

15 cm, what is the charge on the other ball (in

mg

microC)?

Answer:

0.37

Solution: The conditions for static equilibrium on ball Q2 is

KQ1Q2/x

2

KQ1Q2

x2

T cos¦È = mg

T sin ¦È =

and dividing the 1st equation by the second yields

tan ¦È =

KQ1Q2

x 2 mg . Solving for the product of the two charges gives

( 0 .15 m ) 3 ( 2 ¡Á 10 ?3 kg )( 9 .8 m / s 2 )

x 3 mg

=

= 3 .68 ¡Á 10 ?15 C 2

Q1Q 2 =

9

2

2

2 LK

2 (1m )( 8 .99 ¡Á 10 Nm / C )

where I used tan¦È

¦È = x/(2L). Thus,

3.68 ¡Á 10?15 C 2

= 0.368?C

Q2 =

?6

0.01¡Á 10 C

Problem 7:

A charge +Q is fixed two diagonally opposite

corners of a square with sides of length L. On

another corner of the square a charge of +5Q

if fixed as shown in the Figure. On the empty

corner a charge is placed, such that there is no

net electrostatic force acting on the +5Q

charge. What charge is placed on the empty

corner?

Answer:

Solutions

q

+Q

L

+5Q

L

? 2 2Q

Chapter 22

Page 4 of 6

+Q

PHY2049

R. D. Field

Solution: Setting the magnitude of the net for on the +5Q charge due to the

two +Q charges equal to the magnitude of the force on the +5Q charge due

to the charge q gives

2

KQ(5Q) K q (5Q)

=

2 ,

L2

2L

(

)

and solving for the magnitude of q, |q| (and noting that q must be negative)

yields,

q = ?2 2Q .

2Q

Problem 8:

Three point charges Q, 2Q, and 4Q form a

triangle with sides of length L and 2L as

shown in the Figure. What angle does the

resulting electrostatic force on the charge

Q make with the positive x-axis (in

degrees)?

Answer:

7.1

Solution: We see from the figure that

2L

2

2

4KQ /L

4Q

L

¦È

Q

2

F

2 KQ 2 /(2 L) 2 1

tan ¦È =

=

4 KQ 2 / L2

8

and hence ¦È = 7.1o.

+Q

Problem 9:

Two identical point charges +Q are located on

the y-axis at y=+d/2 and y=-d/2 and a third

charge -2Q is located at the origin as shown in

the Figure. The three charged +Q, +Q, and -2Q

form an electrically neutral system called an

electric quadrupole. A charge q is placed on

the x-axis a distance x = d from the quadrupole.

At what is the net force on q due to the quadrupole?

Answer:

Solutions

? 0.57

F2

r

q

x

d

-2Q

¦È

F3

F1

+Q

KQq

x?

2

d

Chapter 22

2

KQ /2L

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