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Question

ABC is a right angled triangle in which AB=3 cm and BC=4 cm. And ABC=π/2. The three charges +15,+12 and 20 e.s.u are placed respectively on A,B and C. The force acting on B is

A
125 dynes
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B
35 dynes
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C
25 dynes
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D
Zero
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Solution

The correct option is C 25 dynes

Let FA be force on B due to A,
FC be force on B due to C,
|FA|=K(15)(12)32=20 dyne
|FC|=K(20)(12)42=15 dyne
Net force on B,
Fnet=F2A+F2C=(202)+(152)
=25 dynes

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