# Two charges of  8 C  and  -3 C are positioned on a line at points  -2  and  -4 , respectively. What is the net force on a charge of  3 C at  1 ?

Dec 19, 2016

$F = 2.076 \times {10}^{10} N$ to the right (positive $x$)

#### Explanation:

The method is to calculate each force (using Coulomb's law) acting on the $3 C$ charge, then do a vector addition of the results to get the net force.

Here it goes:

${F}_{1} = \frac{\left(9 \times {10}^{9}\right) \left(3 C\right) \left(8 C\right)}{3} ^ 2$ = $2.4 \times {10}^{10}$ N

because the $8 C$ and $3 C$ charges are 3 metres (?) apart

${F}_{2} = \frac{\left(9 \times {10}^{9}\right) \left(3 C\right) \left(3 C\right)}{5} ^ 2$ = $3.24 \times {10}^{9}$ N

because the $8 C$ and $3 C$ charges are 3 metres (?) apart

Notice I have deliberately omitted the negative sign on the second $3 C$ charge in ${F}_{2}$. This is because I prefer to use the equation to calculate the magnitude of the force only. I can easily (and more reliably) determine the direction of the force from the layout. If this method is not followed, it is easy to become confused in trying to carry out the vector addition.

${F}_{1}$ is a repulsive force, and acts to the right (positive $x$) on the charge located at 1.
${F}_{2}$ is and attractive force, and acts to the left.

So, the net force is found by subtracting ${F}_{1}$ and ${F}_{2}$ and pointing the resultant to the right (as ${F}_{1}$ is the greater force).

i.e. $F = 2.076 \times {10}^{10} N$ to the right (positive $x$)