# Physics Chapter 16 Solutions Electrostatics Worksheet II

**Pages:**4 (483 words)

**Published:**March 13, 2015

1.

!" F 4.0x10 !17 N

E= =

= 250 N/C

q 1.6x10 !19 C

2.

!"

F = Eq = 75N/C * 2 *1.6x10 !19 C = 2.4x10 !19 N

3.

!" F

0.24N

3

E= =

=

1.2x10

N/C

!4

q 2x10 C

a.

N • m2

* 5x10 !9 C

!" ke q 9x10

2

C

E= 2 =

= 500 N/C

r

(0.3m)2

b.

N • m2

!9

!10

9x10

*

5x10

C

*

4x10

C

2

ke q1 q2

C

F=

=

= 2x10 !7 N

2

2

r

(0.3m)

a.

!" F 2.0x10 !3 N

E= =

= 330 N/C

q

6x10 !6 C

b.

!"

F = Eq = 330N/C * 2.0x10 !6 C = 6.7x10 !4 N

9

4.

9

5.

N • m2

9x10

* 3C

!"

2

ke q

C

E = 2* 2 = 2*

= 2.16x1011 N/C

2

r

(0.5m)

2

6.

7.

a.

!

!E =

ke q1 ke q2

+ 2 =

r2

r

N • m2

C 2 ( q + q ) = 9x10 5 N/C

1

2

(0.5m)2

9x10 9

b.

ke q1

ke q2

= 2

2

r

(r +.1)

(r +.1)

r2

2

q

= 1

q2

r +.1 = 2r

8.

9.

r +.1

q1

2x10 !7 C

=

=

=2

!8

r

q2

5x10 C

r = 0.1m to the right of q 2 . why?

!"

F = mg = Eq = 2x014 N/C*3.2x10 !19 C = 6.4x10 !19 N

Feʼ

T

!

mgʼ

Fr

Fe

mg

" = 30˚ Fe' = mgsin(" )

mg' = mgcos(" )

Fe'

tan(" ) =

so that Fe = mgtan(" )but

mg'

Ke * Q2

Fe =

= mgtan(" ) so that Q =

2

r

mgtan(" )r2

Ke

1x10 #4 kg * 9.8m /s2 * 0.577 * .4m2

substitute in Q =

= 1.0x10 #7 C = .1µC

2

N•m

9.0x109

C2

!

"Pe 5.0x10 #18 J

= 31.3V

10. V = q =

#19

1.6x10

C

11.

a.

N•m

2.0x10 "16 C * 2.0x10 "10 C * 9.0x109

qQKe

C2

Pe =

=

R

.01m

N•m

2.0x10 "10 C * 9.0x109

QKe

C2

V

=

=

b.

R

.01m

!

2

= 3.6x10 "14 J

2

= 180V

$1

1 '

"Pe = Ke * Q1*Q2 * & #

)=

% R1 R2 (

!

12. a.

!

b.

N • m2

#19

9

3C *1.6x10

C * 9.0x10

= 3.24 x10 #9 J

2

C

W 3.24 x10 "9 J

W = Fd therefore F =

=

= 1.08x10 "8 N

d

0.3m

13.

!

a.

!

Ea =

Ke q

R2

Eb =

Ke q

R2

N • m2

9

9.0x10

* 6x10 "6 C

C2

=

= 2.16x105 N /C

(0.5m)2

N • m2

9

9.0x10

* 8x10 "6 C

2

C

=

= 1.28x105 N /C

2

(0.5m)

Mag = E a2 + E b2 = 2.51x105 N /C...

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