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JEE Main Physics Class 12 Electric Charges and Fields Questions
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© examsnet.com
Question : 40
Total: 108
Charges
Q
1
and
Q
2
are at points
A
and
B
of a right angle triangle OAB (see figure). The resultant electric field at point
O
is perpendicular to the hypotenuse, then
Q
1
∕
Q
2
is proportional to :
[Sep. 06,2020 (I)]
x
1
3
x
2
3
x
2
x
1
x
1
x
2
x
2
2
x
1
2
Validate
Solution:
Electric field due charge
Q
2
,
E
2
=
k
Q
2
x
2
2
Electric field due charge
Q
1
,
E
1
=
k
Q
1
x
1
2
From figure,
tan
θ
=
E
2
E
1
=
x
1
x
2
⇒
k
Q
2
x
2
2
×
k
Q
1
x
1
2
=
x
1
x
2
⇒
Q
2
x
1
2
Q
1
x
2
2
=
x
1
x
2
⇒
Q
2
Q
1
=
x
2
x
1
or,
Q
1
Q
2
=
x
1
x
2
© examsnet.com
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