Electric Field Question Help (don't know how to get started)

AI Thread Summary
The discussion focuses on calculating the net electric field E due to two point charges, q1 and q2, separated by a distance of 0.10 m. The key confusion arises from understanding "as a function of x," where x represents various positions along the x-axis, with q1 at the origin. To find the electric field at specific negative x values, one must determine the distances from those points to each charge. The electric field can be calculated using the formula E = kQ/r^2, substituting x for r as needed. Clarification on the concept of x and its application in the calculations is essential for solving the problem.
pierretong
Messages
3
Reaction score
0

Homework Statement





Two point charges q1 = 8.10×10-5 C and q2 = 3.24×10-4 C are separated by a distance d = 0.10 m. Compute their net electric field E (x) as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to the left.

Diagram shows 2 charges q1 and q2 (q1 is on the origin, q2 is distance d to the right along the x axis.

a)What is E(-0.100)?
b)What is E(-0.050)?
c)What is E(-0.010)?

etc...

Homework Equations



E = kQ/r^2

The Attempt at a Solution



no idea where to start this problem. I know how to calculate the electric field but I don't get what it means when it says "as a function of x" (what is x? there is no x on the diagram, just 2 point charges and the distance between them).

If someone could walk me through a, I could do the rest! Thanks!
 
Physics news on Phys.org
pierretong said:
I know how to calculate the electric field but I don't get what it means when it says "as a function of x" (what is x? there is no x on the diagram, just 2 point charges and the distance between them).
x is just the position along the x-axis, measured from q1 (the origin, where x = 0). For example, charge q2 is at x = d = 0.10 m.
 
i don't get it .. do you use the x in place of the r and does q1 stay at the origin? I am really confused on the x concept
 
arl146 said:
i don't get it .. do you use the x in place of the r and does q1 stay at the origin? I am really confused on the x concept
"x" just represents any point along the x-axis. The positions of the charges remain the same.

If you want the field at x = - 0.100, then you have to figure out the distances from that point to the charges in order to compute the field at that point.
 
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...

Similar threads

Back
Top