Electric Field homework problem

AI Thread Summary
The discussion revolves around solving an electric field problem involving three equal charges located in the xy-plane. The user initially calculated the electric field at the origin but arrived at an incorrect magnitude of approximately 32 N/C instead of the correct 20.25 N/C. Key points include the realization that the electric field is a vector quantity, requiring the addition of components rather than treating it as a scalar. The user sought clarification on how to determine the necessary angles for the calculations, ultimately resolving the issue by correctly applying vector components. The discussion highlights the importance of understanding vector addition in electric field problems.
Decan
Messages
43
Reaction score
0
Hello all. I am new to this forum and I'm glad I found this place! Anywho, here's the problem I'm stuck on...

Three equal charges 4.4 micro coulombs are located in the xy-plane, one at (0m, 63m), another at (52m, 0m), and th third at (49m, -53m). Find the magnitude of the electric field at the origin due to these three charges.

Here's how I tried to solve it:

d1 = 63m; d2 = 52m; d3 = 72.18m (derived using the distance formula)

E = kq/d^2
Enet = E1+E2+E3 = kq (1/d1^2+1/d2^2+1/d3^2)

But when I plug the numbers in, I end up with ~32 N/C as my answer but the correct answer is 20.25 N/C. What am I doing wrong? Any help would be greatly appreciated.
 
Physics news on Phys.org
1. Welcome to PF. Notice that we have subforums dedicated to Homework Help. Please use one of those from the next thread on.

2. The electric field is a vector quantity (with a magnitude and a direction in the XY plane), not a scalar. Do you knoow how to add vectors using cartesian (or rectangular) components?
 
I apologize for posting in the wrong sub-forum..thank you for moving it here. As for the problem, I thought about splitting them into components...field 1 does not have an x component and field 2 does not have a y component. For the third field, it does have x and y compoenets, but how do I find the angle needed for the calculation?

So basically, is it:

Ex = kq2/r2^2 + kq3/r3x^2 cos (?)
Ey = kq1/r1^2 + kq3/r3y^2 cos (?)

sqrt(Ex^2 + Ey^2) = E

where ? = mystery angle

I already tried using this...but didnt know what angle to use and the answer I got was close but not exactly to the answer given in the sample problem. Is this correct? If so, any hints on how I can find the mystery angle?
 
Do you really need an angle? You already have the x and y distance components...
 
Figured it out using the angles...thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top