A charge of magnitude q = 4.00 times 10-3 C is placed at each corner of a square

Question

A charge of magnitude q = 4.00 times 10-3 C is placed at each corner of a square with side a = 0.100 m. The charges at the upper left and lower right corners are negative. The charges at the upper right and lower left corners are positive See diagram below. What is the magnitude and direction of the net force on the charge at the lower right corner ? Show all work on this and the next two blank pages. For maximum partial credit, derive your answers using a, q and other symbols until you plug in numbers at the last step. However, do the problem with any method as long as you show your work. Be careful if you plug in numbers earlier in the process. Write neatly and clearly. BOX ANSWERS.

Explanation / Answer

We need to find three force vectors and then add them together. I will do this without numbers until the final step. The general formula to use is F = kqq/d2. Get the three vectors and then add them by vector addition pronciples

Force 1 (upper right to lower right)

F = -kq2/a2 (The negative indicates the direction of force, it is attractive, it pulls upward, positive y)

Force 2 (upper left to lower right)

The distance is the diagonal and is found by the pythagorean theorem

d2 = a2 + a2

d2 = 2a2

F = kq2/2a2 (The positive indicates its repulsive and pushes away at a 45o angle (call it for now)

Therefore this has x and y components found as such

Fx = (kq2/2a2)cos (The direction is a push to the positive x)

Fy = (kq2/2a2)sin   (The direction is a push to the negative y)

Force 3 (lower left to lower right)

F = -kq2/a2 (The negative indicates the direction of force, it is attractive, it pulls left, negative x)

Now add the x components and

Fx = (kq2/2a2)cos - kq2/a2

Simplify

kq2(cos - 2)/2a2   (Net direction is toward negative x)

Now add the y components

Fy = (kq2/2a2)sin - kq2/a2

Simplfy

kq2(sin - 2)/2a2   (Net direction is toward positive y direction)

Now add the x and y together. This will be done using the pythagorean theorem

(Net Force)2 = (kq2(sin - 2)/2a2)2 + (kq2(cos - 2)/2a2)2

Net Force = [(kq2(sin - 2)/2a2)2 + (kq2(cos - 2)/2a2)2]

Now we can put in the numbers

Net Force = [[(9 X 109)(4 X 10-3)2(sin45 - 2)/2(.1)2)]2 + [(9 X 109)(4 X 10-3)2(cos45 - 2)/2(.1)2)]2]

Net Force = 1.32 X 107 N directed at a 45o angle toward the upper left charge. (North of West, so to speak)

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