### You should know¶

#### Coulomb's Law¶

The electrostatics force $\vec{F}$ between two point charges $q_1$ and $q_2$ is

$$ \large \vec{F} = k_e \frac{q_1 q_2}{r^2}\hat{r} $$where

$k_e$: Coulomb's constant $(= \frac{1}{4\pi \epsilon_0})$

$q_{1,2}$: point charges

$r$: distance between the charges.

But in this question, it's not important.

#### Vector Sum¶

Since the electrostatic force is a vector, the net force is the result of adding all the forces together as vectors.

### Solution¶

$$ \large \vec{F}_{d} = k_e \frac{e^2}{d^2} \hat{d} $$$$ \large \vec{F}_{D} = k_e \frac{e^2}{D^2} \hat{D} $$

Thus,

$$ \large \vec{F}_1 = \vec{F}_{d} + \vec{F}_{D} $$$$ \large \vec{F}_2 = \vec{F}_{d} - \vec{F}_{D} $$$$ \large \vec{F}_3 = \sqrt{\vec{F}^2_{d} + \vec{F}^2_{D} }$$Therefore,

$$ \large F_1 > F_3 > F_2 $$### Answer¶

(C) $ F_1 > F_3 > F_2 $