Question

A particle of charge 3.0 x 10 C experiences an upward force of magnitude 4.8 x 10-6N when it is placed in a particular point in an electric field. (Indicate the direction with the signs of your answers. Assume that the positive direction is upward.) (a) What is the electric fleld (in N/C) at that point? N/C (b) If a charge q-1.6 x 10C is placed there, what is the force (in N) on it?

Answer 1

Answer: a) 1.6 * 10 ^-7 N/C (upward) ; b) -2.5*10^-6N (downward)

Explanation: In order to solve this proble we have to use the Coulomb law given by:

F=q*E from this expression we have

E=F/q=4.8*10^-6/30 C= 1.6 * 10 ^-7 N/C

The force on the particle charge by -1.6 X10 C place intead of the initial charge we have

F=q*E= -16 C* 1.6 * 10 ^-7 N/C= -2.5*10^-6N