# Two electric fields perpendicular to each other -- how?

1. Sep 5, 2015

### gracy

1. The problem statement, all variables and given/known data

four point charges are placed at the corners of a square as shown in the figure having side 10 cm.If q is 1μC then what will be electric field intensity at the centre of the square?

2. Relevant equations

3. The attempt at a solution

Half of diagonal
a/√2
0.1/√2
Electric field due to charge "q"
=9 ×10^9 ×1×10^-6/(0.1/√2 )^2=18×10^5
But the thing I m not getting is
In the solution it has been given that at center there are two electric field which are perpendicular to each other.I want to know how?

2. Sep 5, 2015

### blue_leaf77

Don't forget that electric field is a vector quantity, when you want to add several vectors you don't just add their magnitudes, right?
It might refer to the vector components of the electric field at the center.

3. Sep 5, 2015

### gracy

How the figure in question is transformed into the picture above!

4. Sep 5, 2015

### blue_leaf77

Why is your second picture different than the first one? Which one is actually asked to you and what is it precisely that's asked?

5. Sep 5, 2015

### TSny

Maybe they are thinking of using superposition as shown below.

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6. Sep 5, 2015

### gracy

I think We should assume a test charge (positive/negative)at center of square.I have taken positive charge.Now direction of force due to -q and -2q would be opposite and direction of force due to +q and +2g would also be opposite .And electric field would have same direction as force.So resultant direction of force(electric field ) would be directed towards B and A respectively.

Now electric fields are in perpendicular direction.This was all about direction.For magnitude I think the electric field intensity (in second picture )would be because of -q and +q because -2q overcame -q after that -q is left similarly +2q overcame +q and is left with+q .

I don't know whether I am correct,waiting for your corrections of my work.

7. Sep 5, 2015

### blue_leaf77

You don't have to assume a test charge as the force it experiences may be in opposite direction to the actual electric field at that point depending on the test charge's sign.
Your idea is exactly what the superposition principle facilitates, see comment #5.
However, you seemingly encounter a small misconception. It's true that you can assume that the -q part of -2q overcomes the other -q sitting in the opposite corner of the square. But then afterwards, we should have that the contributions from the two confronting -q's disappear. So, what value of charge is actually left and where?

8. Sep 5, 2015

### gracy

No.Think about placing negative charge there.

9. Sep 5, 2015

### blue_leaf77

What does this "no" imply?

10. Sep 5, 2015

### gracy

The electric field vectors will still be perpendicular to each other.

11. Sep 5, 2015

### blue_leaf77

The electric field vectors from each charge?
Yes, they will be.
But that doesn't guarantee the force it experiences to be in the same direction as the electric field at that place. Ok, if you feel easier to assume a test charge, just go this way so long as you keep in mind that the electrostatic force direction is dependent on the charge sign.

12. Sep 5, 2015

### insightful

Sketch the lines of force between the +q and +2q charges with the pairs separated so they don't interact.
Then move them to the square configuration and add more lines of force from +q to -2q, and from -q to +2q.
I would agree that there is only one field at the center, which is the vector sum of two perpendicular fields.

13. Sep 6, 2015

### gracy

yes.if the charge is negative Force and electric field are in opposite directions.This confirms my statement
Both would give the same electric field as in second picture of my #6.

14. Sep 6, 2015

### gracy

15. Sep 6, 2015

### ehild

You need the electric field at the centre of the square. It is the same as the force exerted on a unit positive charge.
The charges at the opposite vertices of the square act along the diagonal, and their force is opposite. So the resultant of the pairs are as drawn in figure 2, they are perpendicular.
I do not follow what you wrote about the magnitude. So what are the magnitude of both forces? And what is their resultant then?

16. Sep 6, 2015

### gracy

But you agree my views on direction part,right?

17. Sep 6, 2015

### gracy

Resultant force F(R) of forces(+2q,and q)would be Directed towards A and Resultant force F'(R) of forces(-2q,and -q) would be Directed towards B.Direction of electric field would be same as of force because I have assumed POSITIVE CHARGE to be at center of square.And magnitudes of electric field would be
E(R)=9 ×10^9 ×2×10^-6/(0.1/√2 )^2-9 ×10^9 ×1×10^-6/(0.1/√2 )^2
=9 ×10^9 ×1×10^-6/(0.1/√2 )^2
And magnitude of E(R)is same as of E'(R).

Last edited: Sep 6, 2015
18. Sep 6, 2015

### ehild

Yes, and what is the resultant, direction and magnitude ?

19. Sep 6, 2015

### gracy

electric field due to negative charge -Q
= - Q/4πϵ0r^2 or Q/4πϵ0r^2?

20. Sep 6, 2015

### ehild

You have shown the direction of the electric field of both the positive and the negative charges in #6. And the magnitude is always non-negative.

In general, the electric field of a point charge q is opposite to that of a point charge -q.

Last edited: Sep 6, 2015