two charges of value 2mC and -50 mC are placed at a distance 80cm apart. . calculate the distance of the point from smaller charge where the intensity will be zero.

Question

Vikas TU · Answer

Amount of first charge is = 2* 10^-6 C Amount of second charge is = 50*10^-6 C Distance = 80 cm Electric Field = kq1q2/r Electric Field will be zero E.F. = 0 (K x 2* 10^-6 *50*10^-6)/80 + ((k*2* 10^-6)/r) = 0 (10^-4/80) + (2x10^-6/r) = 0 10^-4/80 = -(2x10^-6/r) 10^2/80 = 2/r 10/8 = 2/r 10/4 = 1/r R = 0.4

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two charges of value 2mC and -50 mC are placed at a distance 80cm apart. . calculate the distance of the point from smaller charge where the intensity will be zero.

two charges of value 2mC and -50 mC are placed at a distance 80cm apart. . calculate the distance of the point from smaller charge where the intensity will be zero.

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two charges of value 2mC and -50 mC are placed at a distance 80cm apart. . calculate the distance of the point from smaller charge where the intensity will be zero.

two charges of value 2mC and -50 mC are placed at a distance 80cm apart. . calculate the distance of the point from smaller charge where the intensity will be zero.

1 Answers

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Other Related Questions on Electrostatics

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