# Electric Fields and Electric forces help

## Homework Statement

This is the question
Structure opposite comprises 5 ions each of which has lost one electron. The sides of the square formed by the 4 outer ions are of length 0.2nm
Ion E is at the midpoint

1.What is the force on A due to ion A?

## Homework Equations

I'd imagine: F= kq / r^2

## The Attempt at a Solution

I used Pythagoras to get A to E which is 0.141nm which is r
but for the charge would I use 1.6*10^19 even though they've lost an electron?
9*10^9 * 1.6*10*-19 / (0.141*10^-6)^2
I don't think this is right

gneill
Mentor
but for the charge would I use 1.6*10^19 even though they've lost an electron?
Yes, each ion should have a net positive charge equal in magnitude to the charge on an electron (1.6*10^-19 C).
9*10^9 * 1.6*10*-19 / (0.141*10^-6)^2
Careful, what's the order of magnitude of a nanometer?

• IITSRII
9*10^9 * 1.6*10*-19 / (0.141*10^-9)^2 = 7.24*10^10

Yes, each ion should have a net positive charge equal in magnitude to the charge on an electron (1.6*10^-19 C).

Careful, what's the order of magnitude of a nanometer?
How would I find the net electric force on E
I know F= KQq/r^2 but I think I might get confused with the previous answer
not sure where to start from

gneill
Mentor
9*10^9 * 1.6*10*-19 / (0.141*10^-9)^2 = 7.24*10^10
Remember, there are *two* charges involved. Check your formula for the electric force between two charges.
How would I find the net electric force on E
I know F= KQq/r^2 but I think I might get confused with the previous answer
not sure where to start from
Each of the charges forming the square will impart its own force on ##E##. Make a sketch showing the individual forces. What can you say about the size and arrangement of the force vectors?

• IITSRII
Remember, there are *two* charges involved. Check your formula for the electric force between two charges.

Each of the charges forming the square will impart its own force on ##E##. Make a sketch showing the individual forces. What can you say about the size and arrangement of the force vectors?
If theres 2 charges involved wouldn't they have the same charge as they both are ions which lost electrons
9*10^9 * 1.6*10*-19 * 1.6*10*-19 / (0.141*10^-9)^2 =1.15*10^8
since F = KQ1Q2/d^2

For the net electric force on E
would I perhaps sum the individual force of the other 4 ions onto E then combine them? to find out the net electric force on E

gneill
Mentor
9*10^9 * 1.6*10*-19 * 1.6*10*-19 / (0.141*10^-9)^2 =1.15*10^8
Yup. Watch the sign on the power of ten on your result.
For the net electric force on E
would I perhaps sum the individual force of the other 4 ions onto E then combine them? to find out the net electric force on E
Yes. Remember that forces are vector quantities, so the directions count. As I suggested, make a sketch first showing how the force vectors will be arranged. Then, can you solve the problem by inspection?

• IITSRII
Since electric force is KQ1Q2/r2
9*10^9*1.6*10^19*1.6*10^19 / 0.141^2
then multiply that answer by 4 because of the 4 ions
would that be it?

RPinPA
Homework Helper
Since electric force is KQ1Q2/r2
9*10^9*1.6*10^19*1.6*10^19 / 0.141^2
then multiply that answer by 4 because of the 4 ions
would that be it?

No, you're adding scalars. These forces are vectors. Two equal forces can combine to double if they're pulling in the same direction, cancel out if they're pushing in opposite directions, or be somewhere in between if they're acting at an angle relative to each other.

As you've been told, draw the picture. Draw the force from each one, remembering that like charges repel so each corner provides a force pushing directly away from the corner.

Now, look at those four arrows and see if you can draw any conclusions.

No, you're adding scalars. These forces are vectors. Two equal forces can combine to double if they're pulling in the same direction, cancel out if they're pushing in opposite directions, or be somewhere in between if they're acting at an angle relative to each other.

As you've been told, draw the picture. Draw the force from each one, remembering that like charges repel so each corner provides a force pushing directly away from the corner.

Now, look at those four arrows and see if you can draw any conclusions.
already drawn the picture, is it possible that 2 forces repel and 2 forces attract towards E hence 2 forces cancel out
if the structure is square like ABCD with E in the middle of everything

RPinPA
Homework Helper
I thought all the charges here were positive. Which charges do you think are attracting E?

I'm so sorry, I thought the question was asking what the total force on ion E is. I just reread the original post. This is the question you asked.

"1.What is the force on A due to ion A?"

That doesn't make sense. You weren't really asked for the force on A due to itself was, were you? What is it you are actually trying to find?

I thought all the charges here were positive. Which charges do you think are attracting E?

You're on the right track.
ah yeah they are
so negative charges are attracting E since like charges repel
they all cancel out since they are pulling away in opposite directions?

thats a typo
whats the force on E due on ion A

gneill
Mentor
ah yeah they are
so negative charges are attracting E since like charges repel
they all cancel out since they are pulling away in opposite directions?
All the charges are positive since an electron is missing from each. Since all the charges are the same, they all repel each other. However, you are correct in your conclusion that, thanks to the symmetry of the arrangement, all the forces acting on ##E## will cancel.

RPinPA
Homework Helper
thats a typo
whats the force on E due on ion A

The total force on E due to all four other charges is 0, because they come in pairs that cancel each other out.

But if you were just asked about one of those forces, the force on E due to A, it is ##k q_1 q_2/r^2## where ##q_1## and ##q_2## are the charges of A and E (which is the charge from one missing electron) and r is the distance between E and A (which is half the diagonal of the square).

And it points away from A.

• gneill