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Q.

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 net force acting on B is

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An Intiative by Sri Chaitanya

a

125 dyne

b

25 dyne

c

35 dyne

d

Zero

answer is C.

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Detailed Solution

Net force on B is $F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}}$

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$F_{A} = \frac{15 \times 12}{(3)^{2}} = 20 dyne, F_{C} = \frac{12 \times 20}{(4)^{2}} = 15 dyne$

$\Rightarrow F_{net} = \sqrt{F_{A}^{2} + F_{C}^{2}} = \sqrt{(20)^{2} + (15)^{2}} = 25 dyne$

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