Thirty microcoulombs of NEGATIVE charge experiences an electrostatic force of 27. mN. What is the magnitude and direction of the electric field?

To calculate the magnitude and direction of the electric field, we can use the equation:

Electric Field (E) = Force (F) / Charge (q)

Given:
Force (F) = 27 mN = 27 × 10^-3 N (since 1 N = 10^3 mN)
Charge (q) = 30 μC = 30 × 10^-6 C (since 1 C = 10^6 μC)

Let's substitute these values into the equation:

Electric Field (E) = (27 × 10^-3 N) / (30 × 10^-6 C)

Simplifying, we have:

Electric Field (E) = 0.9 N / C = 0.9 V/m (since 1 V/m = 1 N/C)

So, the magnitude of the electric field is 0.9 V/m.

Since the charge quantity is negative, the electric field direction will be in the opposite direction of the force experienced by the charge. In this case, the force is positive, implying that the electric field is directed from positive to negative charge.

Therefore, the direction of the electric field is from the positive charge to the negative charge.