3 point charges each of 5μc are
kept at the vertixes of one equilateral
△ABC of side 20cm. Find the force
on a charge of 2 μC kept at the mid
point of side BC
all charges have same sign
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Answer:
Net electric field at A due to charges at B and C is
E
A
=2E
Ac
sin50
o
=2×9×10
9
×
(0.20)
2
3×10
−6
×
2
3
=
3
×6.75×10
5
AM=
(20)
2
−(10)
2
=
400−100
=10
3
cm
Let the charge at M be q. Charge q should be positive so that there can be repulsion between the charges at A and M.
E
AM
=9×10
9
×
(10
3
/100)
2
q
=3q×10
11
For A to be in equilibrium
E
A
=E
AM
⇒
3
×6.75×10
5
=3q×10
11
or q=
4
9
3
×10
−6
=
4
9
3
μc
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