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Two charges 2 ×10 -8 and -10 -8 are placed at points A and B Respectively. Find the resultant force on a charge of 1 × 10 -6 held at point P, such that AP=10cm, BP=5 cm and Angle APB = 90. [Answer : 4 × 10 -2 N.

Two charges 2 ×10-8 and -10-8 are placed at points A and B Respectively. Find the resultant force on a charge of 1 × 10-6 held at point P, such that AP=10cm, BP=5 cm and  Angle APB = 90. [Answer : 4 × 10-2 N.

Grade:12

1 Answers

Khimraj
3007 Points
2 years ago
force due to 2*10-8 charge
F1 = (9*109)*2*10-8*10-6/(0.1)2 = 1.8*10-2N
force due to -10-8 charge
F2 = (9*109)*10-8*10-6/(0.05)2 = 3.6*10-2 N
angle between F1 and F2 is 90
then resultant is \sqrt{F1^{2} + F2^{2}} = 4*10-2 N

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