question 2

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question 2

Post  shouldbe on Sat Feb 28, 2009 5:11 pm

First, for the love of god, make a picture.

If you do that, you'll figure out that there is definitely an eclectic field.

Where E = kQ/(r^2)

First things first, let's get r.

Since we have a square, and our point of interest is the center of the square. Use a^2 + b^2 = c^2 to get the length of the diagonal. Half that distance is the location of the center point, for all four corners of the square.

Next, figure out an Electric Field for one of the corners, pick one, it doesn't matter. We'll call it 'X'

You should get some large number, don't worry. Next, recall that your only care about the direction of this Electric Field influencing a point in the center, since a square is made up of four right angles, and the point you care about is in the dead center, your diagonal (hypotenuse) where the point lies makes a 45 degree angle with all of the corners.

Since you got the electric field magnitude for one corner, give it a direction now. Which direction? A 45 degree angle.

So take X and multiply it by sin45, make sure you're in degrees you fool.

Great, now you've got your E. Field with a direction, woot.

And I'll leave it to you to figure out how many charges influence the field in this square. Multiply that number by the E.Field(sin45).....(its 4.)

All the charges are the same, therefore the fields they produce are all the same. Cheers.

It's negative because if all the charges were positive, the field would completely cancel itself out. (Draw arrows)

shouldbe
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