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The charge in the bottom right corner of Figure 17.2is Q = - 10 nC. What is the magnitude of the force on Q?

Added Wed, 08 Jul '15

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the force from the top right charge is :

F1 = (9x10^9 x 5x10^-9 x 10x10^-9) / (0.01)^2 = 4.5 x 10^-3 N (downwards)

the force from the top left charge is :

F2 = (9x10^9 x 15x10^-9 x 10x10^-9) / [0.03^2 + 0.01^2] = 1.35 x 10^-3 N (towards 15 nC charge)

the angle between F1 and F2 is :

theta = 180 - tan-1(3/1) = 108.435 degrees.

so F net = sqrt (F1^2 + F2^2 + 2 F1 F2 cos theta )

=> F net = 4.27 x 10^-3 N

Added Wed, 08 Jul '15

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