# Two charges of  2  C  and  4 C are positioned on a line at points  -5  and  -2 , respectively. What is the net force on a charge of  3  C at  1 ?

May 21, 2017

The forces on the $3$ $C$ charge due to each of the other charges can simply be added together.

The net force is $F = \frac{k \times 2 \times 3}{6} ^ 2 + \frac{k \times 4 \times 3}{3} ^ 2 = 1.35 \times {10}^{10}$ $C$

#### Explanation:

The charges in question are all positive, so the forces will be repulsive. The $3$ $C$ charge will be experiencing a force to the right, if we imagine the number line.

The distance from the $2$ $C$ charge to the $3$ $C$ charge is $6$ $m$ (we are not told the units of distance being used, but metres are the appropriate SI unit). The distance from the $4$ $C$ charge to the $3$ $C$ charge is $3$ $m$.

The net force, then, is:

$F = \frac{k \times 2 \times 3}{6} ^ 2 + \frac{k \times 4 \times 3}{3} ^ 2$

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

$F = \frac{k \times 2 \times 3}{6} ^ 2 + \frac{k \times 4 \times 3}{3} ^ 2 = 1.35 \times {10}^{10}$ $C$

This is a very large force, but these are very large charges, quite close together, and the constant is large.