Determine the electric field on the y axis

AI Thread Summary
To determine the electric field on the y-axis at y = 0.470 m due to two 2.04 µC point charges located on the x-axis, the distance from each charge to the point must first be calculated, yielding approximately 1.123 m. The angle between the line connecting the charge and the point on the y-axis is found to be 65.26 degrees. The electric field due to one charge is calculated using E = k*q/r^2, and the y-component is derived from this value. However, since both charges are equal in magnitude and position, their effects must be considered together, leading to potential cancellation of y-components or addition of x-components based on their signs. Properly breaking down the electric field vectors into components and applying the correct algebra is essential for obtaining the accurate result.
LANS
Messages
24
Reaction score
0

Homework Statement


Two 2.04 µC point charges are located on the x axis. One is at x = 1.02 m, and the other is at x = -1.02 m. Determine the electric field on the y-axis at y = 0.470 m.


Homework Equations


E = k*q/r^2


The Attempt at a Solution



First I find the distance from one of the charges to the point on the y-axis.
\sqrt(1.02^2 + 0.47^2) = 1.123m

I also find the angle between this line and the y-axis.
tan^1 \frac{1.02}{0.47} = 65.26

Then I found the component of the electric field at the point on the axis which lies along the y-axis.

E = k_e*\frac{q}{r^2}
E = 8.99e9 * \frac{2.04\mu C}{1.123m}*cos(65.26)

As this only accounts for one charge, and the other charge is equal in magnitude and position, I double this number to get the answer.

Howerver, this gives me the wrong answer. Any help would be appreciated.
 
Physics news on Phys.org
The E field due to one of the charges (the one at x=1.02) on the y-axis is given by your equation E=k*q/r^2.
if you draw a line from your charge to the point on the y-axis, this E vector points in this direction. If your other charge (at x=-1.02) has opposite sign, then the y components of the E vectors cancel each other out and the x-components add. Else, the x-components cancel and the y-components add.

In either case break the E vector that you found into its x and y components using SOH CAH TOA. Then do the simple algebra depending on the signs of your charges.
 
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