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Question

Three equal charges, each having a magnitude of 2.0×106C are placed at the three corners of a right angled triangle of sides 3 cm, 4 cm and 5 cm. The force (in magnitude) on the charge at the right angled corner is :
590134_76a8cb3d9ab84f0e87bca14df24b7289.jpg

A
50 N
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B
26 N
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C
29 N
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D
45.9 N
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Solution

The correct option is D 45.9 N
Charge at each corner q=2.0×106C
Force between two charges F=kq2r2
where r is the distance between them and k=9×109 units.
Force on the charge at the right angled corner |F|=F2x+F2y
Fx=kq2(0.04)2 ^x
Fx=(9×109)×(2.0×106)2(0.04)2=22.5N ^x
Fy=kq2(0.03)2 ^y
Fy=(9×109)×(2.0×106)2(0.03)2=40N ^y
Thus |F|=F2x+F2y
Or |F|=(22.5)2+(40)2
|F|=45.9N

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