Question

What is the magnitude and direction of the electric field at 0.25 m from a -5.6x10^-6 C point charge? Choose the correct answer.

a. 1.18x10⁵⁵ N/C toward the charge

b. 1.18x10⁵ N/C away from the charge

c. 8.10x10⁵ N/C toward the charge

Answer 1

We can find the magnitude of the electric field as follows:

$$E=\dfrac{k|q|}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|-5.6\cdot10^{-6}\ C|}{(0.25\ m)^2}=8.1\cdot10^5\ \dfrac{N}{C}.$$

The electric field directed toward the charge.

Answer:

c. 8.10x10⁵ N/C toward the charge.