What is the net electric field?

Given that ${Q}_{1} = 7 \times {10}^{- 6} C$ is located at the origin and ${Q}_{2} = 5 \times {10}^{- 6} C$ is located $0.3 \text{m}$ to the right of ${Q}_{1}$, what is the net electric field at a point $P$ located $0.4 \text{m}$ above ${Q}_{1}$?

Aug 7, 2017

${E}_{n e t} \approx 1.83 \times {10}^{7} \text{ N"//"C}$

Explanation:

The electric field of a point charge is given by:

$\vec{E} = k \frac{\left\mid q \right\mid}{r} ^ 2$

where $k$ is the electrostatic constant, $q$ is the magnitude of the charge, and $r$ is the radius from the charge to the specified point

The net electric field at point $\text{P}$ is the vector sum of electric fields ${E}_{1}$ and ${E}_{2}$, where:

${\left({E}_{x}\right)}_{n e t} = \sum {E}_{x} = {E}_{x 1} + {E}_{x 2}$

${\left({E}_{y}\right)}_{n e t} = \sum {E}_{y} = {E}_{y 1} + {E}_{y 2}$

${E}_{n e t} = \sqrt{{\left({E}_{x}\right)}^{2} + {\left({E}_{y}\right)}^{2}}$

So, in order to find the net electric field at point $\text{P}$, we will have to analyze the electric field produced by each charge and how they interact (cancel or add together). We can draw a diagram of the situation, keeping in mind that positive charges create electric fields with vectors that point away from them. I will only draw in a couple of the vectors—those that are relevant to the problem—but as in the above picture, the field lines point out (or in) in every direction from the charge.

Diagram: The electric field vector originating from ${Q}_{1}$ which points toward $\text{P}$ has only a perpendicular component, so we will not have to worry about breaking this one up.

Therefore, ${\left({E}_{1}\right)}_{x} = 0$ and ${\left({E}_{1}\right)}_{y} = {E}_{1}$. Since we are given the radius $\left(0.4 \text{m}\right)$, we can calculate ${E}_{1}$:

${E}_{1} = k \frac{\left\mid {Q}_{1} \right\mid}{r} ^ 2$

=((8.99*10^9("N"*"m")/"C"^2)(7*10^-6"C"))/(0.4"m")^2

$= 393312.5 \text{ N"//"C}$

In order to calculate ${E}_{2}$, we will need to find the radius between ${Q}_{2}$ and $\text{P}$. You can see that the electric field vectors of the charges create a right triangle, and since we have both of the side lengths, we can use the Pythagorean theorem to calculate the hypotenuse, our missing radius.

${x}^{2} + {y}^{2} = {r}^{2}$

$\implies r = \sqrt{{x}^{2} + {y}^{2}}$

$= \sqrt{{0.3}^{2} + {0.4}^{2}}$

$= 0.5$

$\therefore$The radius is $0.5 \text{m}$

We can now calculate ${E}_{2}$.

${E}_{2} = k \frac{\left\mid {Q}_{2} \right\mid}{r} ^ 2$

=((8.99*10^9("N"*"m")/"C"^2)(5*10^-6"C"))/(0.5"m")^2

$= 17980000 \text{ N"//"C}$

This field vector occurs at an angle relative to $\text{P}$, however, so we will have to use trigonometry to break it up into its parallel and perpendicular components—just like we do with forces.

We have:

${\left({E}_{2}\right)}_{x} = {E}_{2} \cos \left(\theta\right)$

$\left({E}_{2}\right) y = {E}_{2} \sin \left(\theta\right)$

Before we calculate the components, we'll have to find the angle. We can do this using the arctangent function, since we have both of the triangle's side lengths.

$\tan \left(\theta\right) = \frac{y}{x}$

$\implies \theta = \arctan \left(\frac{y}{x}\right)$

$= \arctan \left(\frac{0.4}{0.3}\right)$

$= {53.13}^{o}$

Therefore:

${\left({E}_{2}\right)}_{x} = \left(17980000 \text{ N"//"C}\right) \cdot \cos \left({53.13}^{o}\right)$

$= 10788025.7 \text{ N"//"C}$

${\left({E}_{2}\right)}_{y} = \left(17980000 \text{ N"//"C}\right) \cdot \sin \left({53.13}^{o}\right)$

$= 14383980.73 \text{ N"//"C}$

Note that because all of these components occur above the positive x-axis and to the right of the origin, all of them have positive values. This may not always be the case, so be sure to keep track of your signs.

We now have:

${E}_{x} = 10788025.7 \text{ N"//"C}$

${E}_{y} = 393312.5 \text{ N"//"C" + 14383980.73" N"//"C}$

$= 14777293.23 \text{ N"//"C}$

We can now find the net electric field at $\text{P}$.

${E}_{n e t} = \sqrt{{\left({E}_{x}\right)}^{2} + {\left({E}_{y}\right)}^{2}}$

$= \sqrt{{\left(10788025.7\right)}^{2} + {\left(14777293.23\right)}^{2}}$

$= 18296171.56 \text{ N"//"C}$

This is a fairly large quantity, so we would likely express it in scientific notation as $\approx 1.83 \times {10}^{7} \text{ N"//"C}$.