# Calculating Electric Force: 2 Electrons & a Proton

• Mdhiggenz
In summary, Homework Equations state that in order to find the resultant of two vectors, one must head-to-tail add the vectors and use the cosine and sine rules to find the direction and magnitude of the resultant.

## Homework Statement

Take a look at this image first

If two electrons are each 2.50×10−10 from a proton, as shown in the figure, find the magnitude and of the net electrical force they will exert on the proton.

## The Attempt at a Solution

k=1/4πε0

So what I did first was calculate the Electric Force in the x directon

Fx= (k*q^2/r^2)cos(theta)

Fy=( k*q^2/r^2)sin(theta)

Then took the magnitude of the sum √(x)^2+(y)^2

And thought that would give me the answer for part one, where did I go wrong?

You seem to have left off the cotribution
The force on the proton is the sum of the forces due to the two electrons.
$$F=\frac{ke^2}{r^2}$$ ... for each electron - but the directions are different.
You also need to show the correct direction for the resultant force.

I don't quite understand? Wouldn't me getting the cos / sin be me taking the directions into consideration?

Mdhiggenz said:
I don't quite understand? Wouldn't me getting the cos / sin be me taking the directions into consideration?

I would suggest you not to find the force in a particular direction. Keep everything in variables. Each electron would exert an equal force in magnitude, say F. The resultant force will be $\sqrt{F^2+F^2+2F^2cosθ}$, where θ=65 degrees. Solve the expression, get it into its simplest form i.e factor out F. Plug in the values.

I got the correct answer using your method, but I don't quite understand how you got that.

It would make sense to me if it was √F^2sin∅+F^2cos

But not the √F^2+f^2+2f^2cos∅

can you elaborate on how you achieved that formula?

Thanks

Mdhiggenz said:
I got the correct answer using your method, but I don't quite understand how you got that.

It would make sense to me if it was √F^2sin∅+F^2cos

But not the √F^2+f^2+2f^2cos∅

can you elaborate on how you achieved that formula?

Thanks

That's the formula for finding resultant of two vectors. Don't you know about it?

Nope never knew about it but I do now (: is it always cos and never sine?

Mdhiggenz said:
Nope never knew about it but I do now (: is it always cos and never sine?

Yes, it is always cosθ. You should check this out from any physics textbook you have. It should be there in a chapter about "Vectors".

I hate to direct people to memorize equations - the formula is a special case of the cosine rule for a scalene triangle.

To add two vectors, draw them head-to-tail (you should have seen this before) ... the resultant goes from the tail of the first one to the head of the second one. This gives you a triangle. Draw it out and you'll see.

You can also use the sine rule for the same triangle in this case because, since it is an isosceles triangle you know all the interior angles. But this is not generally true.

I couldn't find a vector-add diagram online that had all the elements I wanted ... so I drew my own:

... in the above, we want to do ##\vec{F}_{tot}=\vec{F}_A+\vec{F}_B## - magnitudes and directions are shown in the diagram.

In general ##F_A \neq F_B##:

The cosine rule says that: $$F_{tot}^2 = F_A^2+F_B^2-2F_AF_B\cos(A)$$... notice that ##A=180-\phi## (in degrees) and use the identity: ##\cos(180-\phi)=-\cos(\phi)## to get: $$F_{tot}^2 = F_A^2+F_B^2+2F_AF_B\cos(\phi)$$
Once you've got ##F_{tot}## you can use the sine rule to find ##\theta## - the direction of the resultant. $$\frac{F_{tot}}{\sin(A)} = \frac{F_B}{\sin(\theta)} = \frac{F_A}{\sin(180-A-\theta)}$$

This is not normally any more convenient that working with components ... unless...

In the special case where ##F_A=F_B=F## then ##\theta = \phi/2## and the rule becomes: $$F_{tot}^2 = 2F^2\big ( 1+\cos{\phi} \big )$$ ... which can simplify things a great deal.

In general - learn the sine and cosine rules rather than these vector ones, and get used to drawing the diagrams to guide your thinking.

#### Attachments

• vectors.png
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Nice explanation and nice diagram, Simon Bridge!

I was concerned that different countries do things differently: in NZ (and afaik in UK and USA) it is usual for beginning students to go from head-to-tail adding to components without seeing the cosine and sine rules (which will be in a more advanced math course). OP may not be taught from a textbook - and there is no guarantee the textbook will have these relations.

Even with what I did above - they are still mysterious rules that need their own derivation ... which OP will have to look up.

Using components - defining the -x axis as the direction of FA we would write:
$$\vec{F}_A = -F_A\hat{\imath}$$ $$\vec{F}_B=-F_B\cos{\phi}\hat{\imath}-F_B\sin(\phi)\hat{\jmath}$$... then I can do: $$\vec{F}_{tot}=\big (-F_A-F_B\cos{\phi}\big ) \hat{\imath}-F_B\sin(\phi)\hat{\jmath}$$

Then it is a matter of applying Pythagoras to get the total.
It is usually felt that this method is easier for beginners. OP's mistake was leaving one of the forces out.

## What is electric force?

Electric force is a physical phenomenon that describes the attraction or repulsion between two electrically charged particles. It is one of the fundamental forces of nature and is responsible for the behavior of charged particles in the presence of an electric field.

## How is electric force calculated?

Electric force is calculated using Coulomb's law, which states that the magnitude of the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

## What are the units of electric force?

The SI unit of electric force is newton (N). In terms of fundamental units, it can also be expressed as kilogram-meter per second squared (kg⋅m/s²).

## How does the distance between two charged particles affect the electric force between them?

The electric force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance between the particles increases, the force between them decreases.

## What are the factors that affect the electric force between two charged particles?

The electric force between two charged particles is affected by the magnitude of their charges, the distance between them, and the medium in which they exist. It is also affected by the presence of other charged particles in the vicinity.

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