a
Qi Two
charged spheres 10cm apart
attract each other with a force of3.6×10⁵N
What force results from
each of the following charges,considered separately
a)Both charges are doubled and the
distance remains the same.
b)an uncharged, identical sphere is
touched to one of the spheres and then taken far away.
c)the separation is increased to 30cm.
Answers
Answer:
Identify the given information in the problem:
The electric force between the charged spheres "A" and "B" is
F
=
3.0
×
10
−
6
N
if the separation between the charged spheres is
r
=
10
c
m
=
0.10
m
Combined charge:
q
1
+
q
2
=
4
×
10
−
9
C
The magnitude of the electric force between the charges
q
1
and
q
2
can be expressed as:
F
=
k
q
1
q
2
r
2
After plugging in the values, we have:
3.0
×
10
−
6
=
(
9.0
×
10
9
q
1
(
4
×
10
−
9
C
−
q
1
)
)
0.10
2
⇒
(
3.0
×
10
−
6
)
(
0.10
2
)
(
9.0
×
10
9
)
=
(
4
×
10
−
9
)
q
1
−
q
2
1
⇒
1.0
×
10
−
17
3
=
4
×
10
−
9
q
1
−
q
2
1
⇒
1.0
×
10
−
17
=
12
×
10
−
9
q
1
−
3
q
2
1
⇒
3
q
2
1
−
12
×
10
−
9
q
1
+
1.0
×
10
−
17
=
0
By using the quadratic formula, the value of
c
h
a
r
g
e
q
1
can be found as:
q
1
=
−
(
−
12
×
10
−
9
)
±
√
(
−
12
×
10
−
9
)
2
−
4
(
3
)
(
1.0
×
10
−
17
)
2
(
3
)
q
1
=
12
×
10
−
9
±
√
144
×
10
−
18
−
12
×
10
−
17
6
q
1
=
12
×
10
−
9
±
√
24
×
10
−
18
6
q
1
=
12
×
10
−
9
±
4.9
×
10
−
9
6
Thus, the value of the charge
q
1
:
q
1
=
12
×
10
−
9
+
4.9
×
10
−
9
6
=
1.2
×
10
−
9
C
O
r
,
q
1
=
12
×
10
−
9
−
4.9
×
10
−
9
6
=
2.8
×
10
−
9
C
Then,
If the charge
q
1
=
1.2
×
10
−
9
C
, then the value of the charge
q
2
:
q
2
=
4.0
×
10
−
9
−
1.2
×
10
−
9
=
2.8
×
10
−
9
C
Or,
If the charge
q
1
=
1.2
×
10
−
9
C
, then the value of the charge
q
2
:
q
2
=
4.0
×
10
−
9
−
2.8
×
10
−
9
=
1.2
×
10
−
9
C
Part(b).
When charges on both spheres A and B is doubled relative to their original values (everything else is unchanged), then
The electric force between the charged spheres can be expressed as:
F
′
=
k
(
2
q
1
)
(
2
q
2
)
r
2
F
′
=
4
k
q
1
q
2
r
2
Since
k
q
1
q
2
r
2
=
F
, therefore,
F
′
=
4
F
F
′
=
4
(
3.0
×
10
−
6
N
)
F
′
=
12
×
10
−
6
N
Part(c).
When another uncharged sphere C identical to A is touched to B (assuming "A" and "B" are identical spheres ), then half of the charge of the charged sphere "B" would be removed. Hence, the new charge on the sphere "B" would be
q
′
2
=
q
2
/
2
Therefore,
The electric force between the charges charged spheres "A" and "B":
F
′
=
k
q
1
(
q
2
/
2
)
r
2
F
′
=
1
2
(
k
q
1
q
2
r
2
)
Since
k
q
1
q
2
r
2
=
F
, therefore,
F
′
=
1
2
F
F
′
=
1
2
(
3.0
×
10
−
6
N
)
F
′
=
1.5
×
10
−
6
N
Part(d).
If the separation between the charged spheres is
r
′
=
30
c
m
=
0.30
m
, then the electric force between the spheres "A" and "B" can be found as:
F
′
=
k
q
1
q
2
(
r
′
)
2
F
′
=
(
9.0
×
10
9
)
(
1.2
×
10
−
9
C
)
(
2.8
×
10
−
9
C
)
(
0.30
)
2
F
′
=
3.4
×
10
−
7
N