Two charges, -13 C and +5.7 C, are fixed in place and separated by 3.3 m. (a) At

Question

Two charges, -13 C and +5.7 C, are fixed in place and separated by 3.3 m. (a) At what spot along a line through the charges is the net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) (b) What would be the force on a charge of +14 C placed at this spot?

Explanation / Answer

let the charge of -13 C be at x=0 and the charge of 5.7 C be at x= +3.3 m so, electric field at any x = E(x) = sum of field due to the two charges. fox x> 3.3 m (right of 5.7C charge), E(x) = -13/(x^2) +5.7/(x-3.3)^2 for x3.3m) in case two: again solve quadratic equation: 13/(x^2) = 5.7/(x+3.3)^2 gives x= -9.76m, x=-1.98m so at these 3 points, E(x) =0 and force on any charged particle = q*E =0

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