Question

Suppose that a negative charge distribution of in space presents the following expression for electric potential as a function of radius:

V(r)=-k*Q/(r^3-1)

If the charge distribution Q = 1 C. The electrostatic force that a proton would feel (q = 1.6x10-19 C) at a 'r' of 1 x10-9 m distance is:

 

(Use the relationship between electric potential and electric field, followed by the relationship between electric field and electrostatic force)

Response options group

F = + 4.315 x 10 ^ -27 N

F = - 5,124 x 10 ^ -27 N

F = + 5,124 x 10 ^ -27 N

F = - 4,315 x 10 ^ -27 N

F = - 5,124 x 10 ^ -18 N

F = + 5,124 x 10 ^ -18 N

Suppose a system of two charges + Q and -Q at a certain distance 'r' in a vacuum (LaTeX: \ epsilonϵ0), which therefore experience an attractive Coulomb Force (Fe) between them. Suppose further that this two-charge system is subsequently introduced into oil whose dielectric constant (relative permittivity) is LaTeX: \ epsilonϵ = 16 LaTeX: \ epsilonϵ0. How far would you have to place both particles to retain the same attractive force?

Response options

group at half the distance 'r'

a quarter of the distance 'r'

twice the distance 'r'

the interaction distance 'r' does not depend on the dielectric constant of the medium

Answer 1