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9., A stationary oil drop between two parallel plateshas a charge of 3.2 x 10- C and a weight of1.6 x 10 * N. Find the electric field acting on theAns. 5.0 x 10°NCdrop.

Answer»

force = charge * electric field

Here the force acting on the drop is the weight pulling it downwards.

Weight = 1.6*10^-14N

Charge on the drop is 3.2*10^-19C

Hence electric field = (1.6/3.2) * 10^5N/C

= 5*10^4N/C



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