What is the magnitude of an electrostatic force experienced by one elementary charge at a point in an electric field where the field intensity is 3.0 x 10^3 n/c?

a) 1.0 x 10^3 n
b) 1.6 x 10^-19 n
c) 3.0 x 10^3 n
d) 4.8 x 10^-16 n

d) 4.8 x 10^-16 N

Explanation:

The electrostatic force experienced by a charge is given by

where

q is the charge

E is the magnitude of the electric field

In this problem, we have

(elementary charge)

(electric field)

Substituting into the equation, we find the force

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electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j