# Physics Chapter 16 Solutions Electrostatics Worksheet II

Pages: 4 (483 words) Published: March 13, 2015
Solutions Electrostatics Worksheet II

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...