Electric Force and Fields

$F_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}^3} \hat{r}_{12}$

$F_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}^2} \hat{r}_{12}$ (correct)

$F_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}} \hat{r}_{12}$

$F_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r_{12}^4} \hat{r}_{12}$

The force is attractive.

The force is repulsive. (correct)

The force is zero.

The force cannot be determined.

Coulomb (C) (correct)

Ampere (A)

Meter (m)

Newton (N)

<p>For every action, there is an equal and opposite reaction.</p> Signup and view all the answers

<p>8.85 × 10^-12 C^2/N m^2</p> Signup and view all the answers