Recent content by fymdmk

awesome, thanks!

okay so sin25 is definitely in the numerator, I see what you're saying. and I think I've got that ax would be a acostheta and ay would be asintheta. so I add my two vertical vectors, and create a triangle using my combined two vertical vectors tip to toe with the horizontal vector, and find the...

sorry I forgot that the direction is squared, so it would be 0.25 (or 055) ^2 is the sin25 in the numerator then? because it is affecting the charge? that still leaves me with 2 upwards vertical components, and 1 left horizontal component - so then do I make a new triangle to get the total and...

sorry! okay so point A: charge 1 (bottom left) would be: k (7.5e^5C)/0.25m in the upwards direction charge 2 (bottom right) would be split into 2 vectors. 1) find the hypotenuse of the triangle that would get from the charge to point A. this value would be x = sqr rt (50^2 + 25^2) = 55cm =...

right - okay so then its probably my second guess? splitting it into X and Y? so point A would have the Y of the first 75 charge plus the Y of the second 75 charge which would be sin26 0.55 and then the X's would just be the cos26 0.55 of the second 75 charge? does that make sense

Homework Statement In the figure (attached), find the magnitude and direction of the electric field at points A, B, C and D. Homework Equations E = kQ / r^2 The Attempt at a Solution I'm just a little lost with the directions/if I need to be including angles so for point A: E = k 7.5e^5C /...