# Question #e8b64

Jul 5, 2015

The electric field will be zero at 8 cm from the +4Q charge.

#### Explanation:

The first thing you need to know is that the electric field produced by a positive charge always points away from the charge.

This is important to know because it automatically means that the point in which the net electric field is zero must be located between the two charges.

In this case, the fact that one electric field points in one direction and the other in the opposite direction means that you must subtract them to get the net electric field

${E}_{1} - {E}_{2} = 0$

The electric field at a distance $r$ from a charge $Q$ is equal to

$E = K \cdot \frac{Q}{r} ^ 2$

Let's assume that this point is located $x$ centimeters from the $Q$ charge. This means that the distance between this point and the second charge is $12 - x$.

So, this would get you

$\cancel{K} \cdot \frac{\cancel{Q}}{x} ^ 2 - \cancel{K} \cdot \frac{4 \cancel{Q}}{12 - x} ^ 2 = 0$

$\frac{1}{x} ^ 2 - \frac{4}{12 - x} ^ 2 = 0$

Rearrange this equation to get

${\left(12 - x\right)}^{2} - 4 {x}^{2} = 0 \iff - 3 {x}^{2} - 24 x + 144 = 0$

This quadratic equation will produce two values for $x$, a positive one and a negative one. Pick the positive one to get

$x = \text{4 cm}$

So, at 4 cm from the +Q charge the net electric field is zero. The distance from the +4Q charge will thus be

$d = 12 - x = 12 - 4 = \textcolor{g r e e n}{\text{8 cm}}$