Suppose an electric field with a magnitude of 2.8 e 4 n/c at a certain location, points due south. if a charge of -4.0 e -6 c is brought to the location, what is the electrostatic force that acts on this charge?

The correct answer is: F = -0.112N (Negative sign indicates that the force direction is opposite to South: meaning North)

Explanation:
Since,
Electric field = Force / UnitCharge

E = F / q

=> F = Eq

Where F = Electrostatic force = ?
E = Electric field = 2.8 * 10^4 N/C
q = Charge = -4 * 10^-6 C

Plug in the values:
F = (2.8 * 10^4) * (-4 * 10^-6)
F = -0.112N (Negative sign indicates that the force direction is opposite to South: meaning North)

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

One answer could be erosion