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.

Question

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.

Khimraj · Answer

force due to 2*10 -8 charge F1 = (9*10 9 )*2*10 -8 *10 -6 /(0.1) 2 = 1.8*10 -2 N force due to -10 -8 charge F2 = (9*10 9 )*10 -8 *10 -6 /(0.05) 2 = 3.6*10 -2 N angle between F1 and F2 is 90 then resultant is = 4*10 -2 N

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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^-8and -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.

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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.

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Two charges 2 ×10^-8and -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.