# A charge of -5 C is at (-2 , 3 ) and a charge of 3 C is at (8,-2 ) . If both coordinates are in meters, what is the force between the charges?

Jun 23, 2016

$\vec{F} = k \cdot \left(\frac{3}{25}\right) {C}^{2} / {m}^{2} \hat{r}$

#### Explanation:

Firsty, let's look at Coulomb's law:
(1) $\vec{F} = - k \left[\frac{{q}_{\text{1" q_"2}}}{{r}^{2}}\right] \hat{r}$

From the given, we note that the charges are opposite in nature. Thus, the direction of the force is positive, meaning that the charges are attracted to each other.

For this problem, the trickiest part is the coordinates or the distance between the charges. To solve for that, recall the Pythagorean theorem:

(2) ${a}^{2} + {b}^{2} = {c}^{2}$

For our purposes, let's rename $c$ into $r$ along with $a$ into $x$ and $b$ into $y$. Thus,

(3) ${x}^{2} + {y}^{2} = {r}^{2}$

To get $x$, we need to get the sum of theabsolute values of the $x$'s. And same with the $y$'s.

(4) $x = \left\mid {x}_{\text{1") + abs(x_"2}} \right\mid = \left\mid - 2 \right\mid + \left\mid 8 \right\mid = 10$

(5) $y = \left\mid {y}_{\text{1")+abs(y_"2}} \right\mid = \left\mid 3 \right\mid + \left\mid - 2 \right\mid = 5$

Then we just plug in the values from (4) & (5) in (3).

(6)${10}^{2} + {5}^{2} = 100 + 25 = 125 = {r}^{2}$

With that done, just plug in the values of:

• ${q}_{\text{1}} = - 5 C$
• ${q}_{\text{2}} = 3 C$
• $k = 9$x${10}^{9} N \cdot {m}^{2} / {C}^{2}$
• ${r}^{2} = 125$
into Coulomb's Law @(1)