Question

What would be the electric field (magnitude and direction) of 1.50 cm to the right of a charge of -6.5 × 10-6 C?

Answer 1

Answer:

Explanation:

The formula for the electric field is

$E=\frac{kQ}{r^2}$  where E is the magnitude of the electric field, k is Coulomb's constant, Q is the charge of the particle (which is NOT included in the formula), and r is the distance between the centers of the charges (for lack of a better description).

$E=\frac{(9.0*10^9)*(6.5*10^{-6})}{(1.50)^2}$ and we get, to 2 sig fig's

E = 2.6 × 10⁴ to the left (since electric fields are always pointing toward the negative charge and the electric field is to the right of the negative charge)