# Two charges of  -5  C  and  -3  C are positioned on a line at points  -8  and  6 , respectively. What is the net force on a charge of  3 C at  -1 ?

Mar 9, 2016

This is a linear problem, so we can just add up the forces on the $3$ $C$ charge due to each of the others. The resultant force is $1.11$ $N$ to the left.

#### Explanation:

The forces on particles with charges ${q}_{1}$ and ${q}_{2}$ a distance $r$ $m$ apart is given by:

$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

Where $k = 9 \times {10}^{9}$ $N {m}^{2} {C}^{-} 2$

To calculate the force on the $3$ $C$ charge due to the $- 5$ $C$ charge, note that, at $- 8$, it is $7$ $m$ away from an object at $- 1$.

${F}_{1} = \frac{k \left(- 5\right) \left(3\right)}{7} ^ 2 = - 2.76$ $N$ (substituting in the value of $k$)

Since the charges are of opposite signs, this is an attractive force, acting to the left (i.e. the negative direction on the number line).

To calculate the force due to the $- 3$ $C$ charge, note that at $6$ it is also $7$ $m$ away from an object at $- 1$.

${F}_{1} = \frac{k \left(- 3\right) \left(3\right)}{7} ^ 2 = - 1.65$ $N$

This is also an attractive force, due to the opposite charges, but it is acting to the right.

Subtracting $1.65$ from $2.76$ we find that there is a net force on the $3$ $C$ charge of $1.11$ $N$ to the left.