Question

14. Charges 25Q, 9Q and Q are placed at points
ABC such that AB = 4m, BC = 3m and
angle between AB and BC is 90°. Then
force on the charge C is

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Answer 1

see figure, a triangle ABC is drawn in such a way that, AB = 4m, BC = 3m and angle ABC = 90° and also point A contains 25Q charges , B contains 9Q and C contains Q charge.

now, force due to point B on point C , F1= K(9Q)(Q)/(3)²

= 9KQ²/9 = KQ² .......(1)

force due to point A on point C , F2= K(25Q)(Q)/(5)²

= 25KQ²/25 = KQ².......(2)

here it is clear that F1 = F2 = KQ² = F (assumed)

now, net force on point C = √(F1² + F2² + 2F1F2cosBCA}

cosBCA = BC/AC = 3/5

so, net force on point C = √(F² + F² + 2F² × 3/5)

= √(2F² + 1.2F²) = √(3.2)F

hence, force on charge C is √(3.2)KQ²