Three point charges of +2C, - 3C and - 3C are kept at
the vertices A, B and C respectively of an equilateral
triangle of side 20 cm as shown in the figure. What should
be the sign and magnitude of the charge to be placed at the
mid-point (M) of side BC so that the charge at A remains in
equilibrium?
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Answer:
Explanation:
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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