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AP and TS EAMCET Engineering Model Paper 14
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© examsnet.com
Question : 108
Total: 159
Force between two indential charges placed at a distance r in vaccum is F. Now a slab of diectric constant 4 and thickness
r
3
is inserted between these two charges. The new force between the charges is
9
F
16
9
F
4
F
3
F
4
Validate
Solution:
F
=
1
4
π
∈
0
q
1
q
2
[
√
k
1
r
1
+
√
k
2
r
2
]
2
=
1
4
π
∈
0
q
1
q
2
[
√
4
×
r
3
+
1
×
2
r
3
]
2
=
1
4
π
∈
0
q
1
q
2
(
4
r
3
)
2
=
9
16
F
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