Two tiny objects with equal charges of 6.59 µC are placed at the two lower corne
ID: 1705224 • Letter: T
Question
Two tiny objects with equal charges of 6.59 µC are placed at the two lower corners of a square with sides of 0.500 m. Find the electric field at the point midway between the upper left and right corners.
So I drew a square and the bottom two corners have the 6.59 µC charge. We want to find the electric field at a point that is located at the top middle of the square. The distance from either of the charges to the point (lets call it A) is sqrt(0.5^2+0.25^2)= 0.559 m.
The formula for the electric field is (k*q)/r^2 where k=8.988e9, q=6.59e-6 and r^2=0.3125. So plug this into the formula and you get 189538.944. But we have 2 charges and they have the same magnitude so you multiply this answer by 2 and get 379077.888 N/C. BUT this answer isn't working so I am obviously doing something wrong. Any suggestions?
Thanks!
Explanation / Answer
You are not far off from the answer. But you assumed that the magnitutdes were in the same direction, but if you look at the diagram, the horizontal magnitudes cancel out, and the magnitude should be pointing straight down. The easiest way to do this is using proportions. You already found the distance from a charge to point A and you know the vertical distance from point A straight down, which is .5 m. so you can do this: .559/.5 (diagonal over vertical) = 189538.944/x (diagonal magnitude/vertical magnitude) x should be 158212 N/C, which is the veritcal of the magnitude of one charge. All you have to do is multiply by 2 and that should be the answer, 316425.616 N/C straight down.
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