Solving the Electric Field at Center of Square: Magnitude & Direction

[GenMipps]

This homework assignment is killing me. I keep thinking I know what I'm doing, and I keep getting it wrong. If anyone can tell me how I goofed this one up, I'd appreciate it.

What they give in the figure is a square. Upper left is +q, upper right is -2q, bottom right is +2q, and bottom left is -q. Each side of the square is given as a. Then the question reads:

What are the magnitude and direction of the electric field at the center of the square of Fig. 23-30 if q= 2.21 x10-8 C and a= 4.25 cm ?

Now, What I did was drew myself a diagram of this, and rotated it 45 degrees clockwise, so E3 was pointing up, E2 was pointing to the right, E1 was pointing down, and E4 was pointing to the left. So, using E = kq/r2, I found the electric fields. R has to be 3 using pythagoram's theorem (4.25^2+4.25^2 then divide by 2).

So, I got E1 = 22.0754 N/c, E2 = -44.151, E3 = 44.151 and E4 = -22.054. Now, I drew a diagram with these, and found the horizontal components (E2 and E4) were summed together to 22.0756. The vertical components were summed together and found to be -22.0756. So, with x and y components, the resultant should be 31.22. I then subtracted 45 degrees again since I had rotated it at the beginning. That gave the direction in the vertical direction.

However, when I entered this answer in, I got the magnitude wrong, but the direction correct. Where did I slip up?

 

[Farsight]

This just shows what I don't know. I thought the answer was 0. Sorry.

 

[nrqed]

This homework assignment is killing me. I keep thinking I know what I'm doing, and I keep getting it wrong. If anyone can tell me how I goofed this one up, I'd appreciate it.

What they give in the figure is a square. Upper left is +q, upper right is -2q, bottom right is +2q, and bottom left is -q. Each side of the square is given as a. Then the question reads:

What are the magnitude and direction of the electric field at the center of the square of Fig. 23-30 if q= 2.21 x10-8 C and a= 4.25 cm ?

Now, What I did was drew myself a diagram of this, and rotated it 45 degrees clockwise, so E3 was pointing up, E2 was pointing to the right, E1 was pointing down, and E4 was pointing to the left. So, using E = kq/r2, I found the electric fields. R has to be 3 using pythagoram's theorem (4.25^2+4.25^2 then divide by 2).

So, I got E1 = 22.0754 N/c, E2 = -44.151, E3 = 44.151 and E4 = -22.054. Now, I drew a diagram with these, and found the horizontal components (E2 and E4) were summed together to 22.0756. The vertical components were summed together and found to be -22.0756. So, with x and y components, the resultant should be 31.22. I then subtracted 45 degrees again since I had rotated it at the beginning. That gave the direction in the vertical direction.

However, when I entered this answer in, I got the magnitude wrong, but the direction correct. Where did I slip up?

You must put the distance in *meters* in order to get the E field in N/C.

Patrick

 

[nrqed]

This homework assignment is killing me. I keep thinking I know what I'm doing, and I keep getting it wrong. If anyone can tell me how I goofed this one up, I'd appreciate it.

What they give in the figure is a square. Upper left is +q, upper right is -2q, bottom right is +2q, and bottom left is -q. Each side of the square is given as a. Then the question reads:

What are the magnitude and direction of the electric field at the center of the square of Fig. 23-30 if q= 2.21 x10-8 C and a= 4.25 cm ?

Now, What I did was drew myself a diagram of this, and rotated it 45 degrees clockwise, so E3 was pointing up, E2 was pointing to the right, E1 was pointing down, and E4 was pointing to the left. So, using E = kq/r2, I found the electric fields. R has to be 3 using pythagoram's theorem (4.25^2+4.25^2 then divide by 2).

So, I got E1 = 22.0754 N/c, E2 = -44.151, E3 = 44.151 and E4 = -22.054. Now, I drew a diagram with these, and found the horizontal components (E2 and E4) were summed together to 22.0756. The vertical components were summed together and found to be -22.0756. So, with x and y components, the resultant should be 31.22. I then subtracted 45 degrees again since I had rotated it at the beginning. That gave the direction in the vertical direction.

However, when I entered this answer in, I got the magnitude wrong, but the direction correct. Where did I slip up?

You must put the distance in *meters* in order to get the E field in N/C.

Patrick