Physics Help Forum Magnitude and Direction of Charge
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 Jan 24th 2010, 03:45 PM #1 Junior Member   Join Date: Jan 2010 Posts: 2 Magnitude and Direction of Charge Four point charges are at the corners of a square of side a as shown in the attached figure. (A=5, B=4, C=4) a.) Determine the magnitude and direction of the electric field at the location of charge q (use k_e for the Coulomb constant and q and a as necessary) b.) What is the resultant force on q? in this problem, i found each of the vectors first. Through the work I did, I came to an answer of (6.77 z + 5.77 j) k_eq/a^2. from there, I found the square root of [(6.77)^2 + (5.77)^2] k_eq/a^2, yielding 8.895 k_eq/a^2. the above answer ended up being wrong, but the angle that i got from it (using the arctangent of 6.77 and 5.77 ) was correct (40.44 degrees) help! what is the correct magnitude and where did i go wrong?
 Jan 24th 2010, 03:48 PM #2 Junior Member   Join Date: Jan 2010 Posts: 2 the image. here is the image for the problem... sorry im new at this. Attached Images
Jan 25th 2010, 08:34 AM   #3
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Hmmm...

 Originally Posted by scotchtapesticky Four point charges are at the corners of a square of side a as shown in the attached figure. (A=5, B=4, C=4) a.) Determine the magnitude and direction of the electric field at the location of charge q (use k_e for the Coulomb constant and q and a as necessary) b.) What is the resultant force on q? in this problem, i found each of the vectors first. Through the work I did, I came to an answer of (6.77 z + 5.77 j) k_eq/a^2. from there, I found the square root of [(6.77)^2 + (5.77)^2] k_eq/a^2, yielding 8.895 k_eq/a^2. the above answer ended up being wrong, but the angle that i got from it (using the arctangent of 6.77 and 5.77 ) was correct (40.44 degrees) help! what is the correct magnitude and where did i go wrong?
Hi there,
Let's go through and find the E field due to each point charge seen at point charge q. If we let q be at the origin then we have three different vectors that need to be summed. Let's start with the E field due to point Aq:

E(Aq) = -(5kq/a^2)i where k = Coulomb constant = 1/(4*pi*permittivity of free space) and i is the unit vector pointing along the x axis.

Now for the charge at point Cq:

E(Cq) = -(4kq/a^2)j where j is the unit vector for the y axis

Now for the charge at point Bq:

Since point Bq is really just a distance away from point q by the hypotenuse of a right triangle with sides of a, then the distance from q to point Bq must just be sqrt(a^2 +a^2) = sqrt(2)*a, and when placed into the E field function it will be squared yielding 2a^2.

E(Bq) = cos(225)4kq/(2a^2)i +sin(225)4kq/(2a^2)j
E(Bq) = (-1/sqrt(2))*(2kq/a^2)i + (-1/sqrt(2))*(2kq/a^2)j
E(Bq) = (-sqrt(2)kq/a^2)i + (-sqrt(2)kq/a^2)j

The reason why the angle is 225 is becasue point Bq is really 45 degrees below the negative x axis with our charge q at the origin. So the angle of point Bq with respect to point q is 180 + 45 degrees around the origin or 225 with respect to the positive x axis.

Now summing all the E fields from each point charge yields:

Etotal = -(5kq/a^2)i - (sqrt(2)kq/a^2)i - (4kq/a^2)j - (sqrt(2)kq/a^2)j

E = -[(5kq + sqrt(2)kq)/(a^2)]i + [-(4kq + sqrt(2)kq)/(a^2)]j

E = [-(5 + sqrt(2))i - (4 + sqrt(2))j](kq/a^2)

The magnitude would be simply the square root of the sum of the squared components:

|E| = (8.393798)(kq/a^2)

and the angle would be the arctan of the imaginary part over the real part:

Angle = arctan[(4 + sqrt(2))/(5 + sqrt(2))] = 40.167586 degrees

And remember that we took q to be at the origin, so the direction of the E field as can be seen from the fact that the two components lay within the third quadrant (they are both negative) means the vector points down into the 3rd quadrant. So whatever angle you get by taking the arctan of the expression above should be added to 180 degrees to put it into the third quadrant. Much like when we had to add 180 degrees to the 45 degree angle at which point Bq is located with respect to the negative x axis, or 225 degrees with respect to the positive x axis. So the actual angle given our convention that we chose is 220.167586 degrees.

As far as the force acting on charge q due to this E field that is simply the product of the charge q and the E field, so the force vector acting on charge q is:

F = [-(5 + sqrt(2))i - (4 + sqrt(2))j]k(q/a)^2

I hope that helps.

Many Smiles,
Craig

Last edited by clombard1973; Jan 25th 2010 at 08:59 AM.

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