Electric Field Strength magnitude

In summary, four point charges are placed at the corners of a square, with two positive charges on the top corners and two negative charges on the bottom corners. At the center of the square, the magnitude of the electric field strength due to one of the charges is (kQ)/2a2. While the charges at the corners have the same magnitude, they have different directions, resulting in a net electric field at the center point, which can be seen by drawing a diagram.
  • #1
Peter G.
442
0
Hi,

Four point charges of equal magnitude are held at the corners of a square. The length of each side of the square is 2a and the sign of the charges are the following: Two positive charges on the top two corners of the square and two negative charges on the bottom two corners of the square.

The point P is at the center of the square.

a) Deduce that the magnitude of the electric field strength at Point P due to one of the point charges is equal to (kQ)/2a2:

I managed to do this one by proving that r2, that is, the distance between the center of the charge and the point P is equal to 2a2 by doing Pythagoras

The next two parts of the question however, ask me to draw an arrow showing the resultant force and then determine, in terms of Q, a and k the magnitude of the electric field at point P. I simply can't do it however because it seems to me that there will be no resultant electric field at P! There are two positive charges and two negative charges, all at the same distance from point P and all have the same magnitude!

Can anyone help me with this (but don't give out the answer, I want to to try and work it out :smile:)

Thanks,
Peter
 
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  • #2
There are two positive charges and two negative charges, all at the same distance from point P and all have the same magnitude!
But different directions - don't forget the sign of the charges.
 
  • #3
But if they have opposite signs they have opposite directions and that's why I thought they would cancel. Or should I assume point P is a positive test charge or something?
 
  • #4
Electric field strength means the force that a unit + charge would experience so you are very much on the right track.
The unit of field strength is N/C
draw a diagram showing the + charges and the - charges
Draw the arrows showing the forces on a + charge at the centre of the square.
Can you see the resultant?
 
  • #5
If they were all the same sign, then opposite corners would cancel out.Draw the situation for a positive charge at P.
The charge wants to move towards -ve charges and away from +ve charges.
See?

[edit] I was beaten to it:
+1 the tech person: always draw the picture.
 
  • #6
very gracious of you to say that Simon !
Cheers
 
  • #7
Thanks a lot for the contribution guys! I got it now. From what I drew I believe there would be a resultant downward electric field. Thanks!
 

1. What is the unit for electric field strength magnitude?

The unit for electric field strength magnitude is Newtons per Coulomb (N/C).

2. How is electric field strength magnitude calculated?

Electric field strength magnitude is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. It can also be calculated by dividing the potential difference by the distance between the charges.

3. What factors affect the magnitude of an electric field?

The magnitude of an electric field is affected by the distance between the charges, the magnitude of the charges, and the medium through which the charges are separated.

4. What is the difference between electric field strength and electric field intensity?

Electric field strength and electric field intensity are used interchangeably and refer to the same concept. Both terms describe the magnitude of the electric field at a given point.

5. How does the direction of the electric field affect its magnitude?

The direction of the electric field does not affect its magnitude. The magnitude of the electric field is only dependent on the distance between the charges and the magnitude of the charges.

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