# A point charge of 3.00×10^-6 C is 12.0cm distant from a second point charge of −1.50×10^-6 C. Calculate the magnitude of the force on each charge?

May 18, 2017

$2.813 \text{ N}$

#### Explanation:

Write out all the givens for this problem.

$\text{Given}$
q_(1) = 3.00xx10^(−6)C
${q}_{2} = - 1.50 \times {10}^{- 6} C$
r = 12.0" cm"or(0.12" m")

To calculate the Coulombic Force between 2 point charges, you use the following:

color(white)(aaaaaaaaaaaaaaaa)color(magenta)(F_(c) = (k |q_(1)|q_(2)|)/r^2

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\text{Since we are only looking for the}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\text{magnitude of the force, we use}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\text{the absolute values of each point}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\text{charge, so the signs are ignored}$

Where
k = "Coulomb's constant" (9xx10^9" (N*m^2)/(C^2))

$- - - - - - - - - - - - - - - - - - - - -$

Plugin

color(magenta)(F_(c) = (k |q_(1)|q_(2)|)/r^2

F_(c) = [(9*10^9" (N*cancelm^2)/(cancel(C^2))) * (3.00*10^(−6)cancelC)*(1.50*10^(-6)cancelC)]/(0.12cancelm)^2

color(blue)(|F_(c)|= 2.813" N"

Note: If we included the signs in the formula, a (-) force would indicate attraction between the two charges and a (+) force would indicate a repulsion.

$\textcolor{w h i t e}{a a a}$

The force on ${q}_{1}$ due to ${q}_{2}$ is $2.813 \text{ N}$ and the force on ${q}_{2}$ due to ${q}_{1}$ is also $2.813 \text{ N}$. Magnitude of the force is the same but the direction is opposite.

$\textcolor{w h i t e}{a a a a a a a a a}$