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

125

dynes

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35

dynes

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25

dynes

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Zero

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Solution

The correct option is C $25$ dynes

Let $F_{A}$ be force on $B$ due to $A,$
$F_{C}$ be force on $B$ due to $C,$
$| F_{A} | = \frac{K (15) (12)}{3^{2}} = 20$ dyne
$| F_{C} | = \frac{K (20) (12)}{4^{2}} = 15$ dyne
Net force on B,
$F_{n e t} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20^{2}) + (15^{2})}$
$= 25$ dynes

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