# Electric field strength in a square.

1. Jun 4, 2006

### al_201314

Hi guys,

Suppose I have a point O in the centre of the square (of side length X), where the top 2 corners of the square carries a positive charge each and the bottom left corner a negative charge and the bottom right corner a positive charge, what would the magnitude and direction of the magnetic field strength at point O be? I figured that it would be towards M, but how do I find it's magnitude?

The given answer is q/(pi)(Enot)(x^2). I don't realy understand why is the 4 missing?

Thanks!

2. Jun 4, 2006

### siddharth

First find the electric field due to each point charge, and then find the vector sum.
One more thing, you do mean electric field, don't you? Why would there be a magnetic field?

3. Jun 4, 2006

### al_201314

Thanks for the help. I apologise I did mean electric field was thinking of some other question at the same time!

4. Jun 4, 2006

### al_201314

Hi Siddarth

Wodner if you could help me out further here.. after some doing I realised I couldn't get rid of the 4, but it would instead be 2.

Here's my working, the electric field strength due to the top left and bottom right corner would be 0 as they cancel out (am I correct?). But the field strength due to the top right and bottom left corner would be 2q/4(pi)(Enot)(x^2)?

Thanks

5. Jun 4, 2006

### siddharth

Yeah, if the charges are the same.

Not quite, what's the distance of the center of the square O from each vertex? (It isn't x)

Also, I don't get the given answer. Are you sure the charges are equal?

6. Jun 4, 2006

### al_201314

Thanks Siddarth for the assistance.. I think I got the answer..

Yes the question states that 4 equal charges except for the bottom left is a negatively charged one compared to the rest of the positively charged ones.

Here's my complete working, I'll name the sides J, K L and M for ident purposes starting from the top left corner in a clockwise direction. So M is the point that carries the -ve charge. E field due J and L cancels each other. The (distance)^2 between K and O I calculated to be (x^2)/2 which is also the (distance)^2 between O and M. So by vector addition, E field due K and M should be in the same direction I got [q/4(pi)(Enot)(x^2)/2] + [q/4(pi)(Enot)(x^2)/2] which gives the answer to be q/(pi)(Enot0)(x^2) after cancelling out the numbers.

Thanks once again.

7. Jun 4, 2006

### siddharth

Yeah, that's right.