Apply Coulomb's Law
Physics · Flashcards · Application
PHYSICS · FLASHCARDS
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QUESTION
Calculate
F
net
F_{\text{net}}
F
net
on
q
2
q_2
q
2
if
q
1
=
+
1
×
10
−
6
C
q_1=+1\times10^{-6}\ \text{C}
q
1
=
+
1
×
1
0
−
6
C
at
0.10
m
0.10\ \text{m}
0.10
m
left and
q
3
=
+
1
×
10
−
6
C
q_3=+1\times10^{-6}\ \text{C}
q
3
=
+
1
×
1
0
−
6
C
at
0.10
m
0.10\ \text{m}
0.10
m
right.
QUESTION
Calculate
F
F
F
for
q
1
=
+
1
×
10
−
6
C
q_1=+1\times 10^{-6}\ \text{C}
q
1
=
+
1
×
1
0
−
6
C
,
q
2
=
−
1
×
10
−
6
C
q_2=-1\times 10^{-6}\ \text{C}
q
2
=
−
1
×
1
0
−
6
C
,
r
=
0.10
m
r=0.10\ \text{m}
r
=
0.10
m
.
QUESTION
Find
q
2
q_2
q
2
if
F
=
0.90
N
F=0.90\ \text{N}
F
=
0.90
N
,
q
1
=
+
1
×
10
−
6
C
q_1=+1\times 10^{-6}\ \text{C}
q
1
=
+
1
×
1
0
−
6
C
, and
r
=
0.10
m
r=0.10\ \text{m}
r
=
0.10
m
.
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ANSWER
∣
q
2
∣
≈
1.0
×
10
−
6
C
|q_2|\approx 1.0\times 10^{-6}\ \text{C}
∣
q
2
∣
≈
1.0
×
1
0
−
6
C
.
∣
q
2
∣
=
F
r
2
k
∣
q
1
∣
=
0.90
×
0.01
8.99
×
10
9
×
10
−
6
≈
1.0
×
10
−
6
C
|q_2| = \frac{Fr^2}{k|q_1|} = \frac{0.90 \times 0.01}{8.99\times10^9 \times 10^{-6}} \approx 1.0\times10^{-6}\ \text{C}
∣
q
2
∣
=
k
∣
q
1
∣
F
r
2
=
8.99
×
1
0
9
×
1
0
−
6
0.90
×
0.01
≈
1.0
×
1
0
−
6
C
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