# A point particle with charge -3q (where q = 420 nC) is located at the origin of a coordinate...

## Question:

A point particle with charge -3q (where q = 420 *nC*) is located at the origin of a coordinate system. Another point particle with charge +q is located at x = 32.0 *cm*. Find the location (in cm) of a point along the x-axis where the electric field is zero, which is not between the two charges.

## Coulomb's law:

When two charges are placed in a space, then there is some interaction between these two charged particles which makes them to attract or to repel each other. This intersection is called the coulomb's force, defined as:

{eq}\begin{align*} \vec F = \dfrac{kq_1q_2}{|\vec r |^2} \hat r \end{align*} {/eq}

where {eq}q_1 {/eq} and {eq}q_2 {/eq} are the charges and should be put with the sign of the charge, {eq}\vec r {/eq} is the displacement vector between the two charges, and {eq}k = 9\times 10^9 {/eq} is the constant.

We see that there is some force of interaction, but what is the reason behind this force?

A charged particle forms an electric field in the surrounding which when interacting with the other charged particles imposes the force of interaction on the other charge. This electric field can be calculated as:

{eq}\begin{align*} \vec E = \dfrac{kq}{|\vec r |^2} \hat r \end{align*} {/eq}

Here, the q is the charge whose electric field we wish to calculate.

## Answer and Explanation:

We are given the two charges -3q (q = 420 nC) and +q located at the position x = 0 and x = 32 cm = 0.32 m respectively.

Let us take the position 'x m' on the x-axis where we want to calculate the electric field. We will first calculate the individual electric fields due to both the charges and then we'll vectorially add them to get the net electric field.

The electric field due to -3q charge at the position x will be:

(The displacement vector from the origin to the x will be{eq}\implies \vec r _1 = x \hat i {/eq})

$$\begin{align*} \vec E _1 &= \dfrac{kq_1}{|\vec r _1|^2}\hat r_1 \\ \vec E _1 &= \dfrac{(9\times 10^9)(-3\times 420\times 10^{-9})}{x^2}\hat i \\ \vec E _1 &= -\dfrac{9\times 1260}{x^2}\hat i\\ \vec E _1 &= - \dfrac{11340}{x^2}\hat i \\ \end{align*} $$

And the electric filed due the charge q will be:

(The displacement vector from the position x = 0.32 m to the x will be {eq}\implies \vec r _2 = (x - 0.32 )\hat i {/eq})

$$\begin{align*} \vec E _2 &= \dfrac{kq_2}{|\vec r _2|^2}\hat r_2 \\ \vec E _2 &= \dfrac{(9\times 10^9)(420\times 10^{-9})}{(x-0.32)^2}\hat i \\ \vec E _2 &= \dfrac{3780}{(x-0.32)^2}\hat i \\ \end{align*} $$

The net electric field at the position x is zero, so we do vector addition of the fields and set equal to zero.

$$\begin{align*} \vec{E}_1 + \vec{E}_2 &= 0 \\ - \dfrac{11340}{x^2}\hat i + \dfrac{3780}{(x-0.32)^2}\hat i &= 0 \\ \dfrac{3780}{(x-0.32)^2} &= \dfrac{11340}{x^2} \\ \dfrac{1}{x^2-0.64x + 0.1024} &= \dfrac{3}{x^2}\\ x^2 &= 3x^2 - 1.92x + 0.3072 \\ 2x^2 - 1.92x + 0.3072 &= 0 \\ \implies x &= 0.7571 \ m, 0.202 \ m \end{align*} $$

The only possible solution is **x = 0.7571 m **.Thus at **x = 75.71 cm**, the net electric field is zero.