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KCET 2015 Physics Solved Paper
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© examsnet.com
Question : 16
Total: 60
Two spheres carrying charges +6µC and +9µC. separated by a distance d, experiences a force of repulsion F. When a charge of -3µC is given to both the sphere and kept at the same distance as before, the new force of repulsion is
F
3F
F/3
F/9
Validate
Solution:
Given, charge
q
1
=
+
6
µ
C
=
6
×
10
−
6
C
;
charge
q
2
=
+
9
µ
C
=
9
×
10
−
6
C
seperation = d
Force between charges is given as
F
=
1
4
π
ε
0
q
1
q
2
d
2
∝
q
1
q
2
d
2
=
6
×
10
−
6
×
9
×
10
−
6
d
2
→ (1)
When a charge of +3 μC is given to both and separation = d, then
q
1
′
=
+
6
µ
C
−
3
µ
C
=
3
µ
C
=
3
×
10
−
6
C
q
2
′
=
+
9
µ
C
−
3
µ
C
=
6
µ
C
=
6
×
10
−
6
C
The force between charges is given as
F
′
=
1
4
π
ε
0
q
′
1
q
′
2
d
2
∝
q
′
1
q
′
2
d
2
=
3
×
10
−
6
×
6
×
10
−
6
d
2
→ (2)
Dividing Eq. (1) by Eq. (2), we get
F
F
′
=
6
×
10
−
6
×
9
×
10
−
6
3
×
10
−
6
×
6
×
10
−
6
=
3
⇒
F
′
=
F
3
Therefore, the new force of repulsion is
F
3
© examsnet.com
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