Electric field at the origin of a quarter-circle

  • Thread starter henry3369
  • Start date
  • #1
194
0

Homework Statement


Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of curvature at the origin. Find the x- and y-components of the net electric field at the origin.

Homework Equations


V = ∫kdQ/r
Eradial = -dV/dr

The Attempt at a Solution


V = ∫kdQ/r = -kQ/r
Eradial = -dV/dr = -(d/dr)(-kQ/r) = kQ(d/dr)(1/r) = kQ(-1/r2) = -kQ/r2 = -kQ/a2.

I don't know how to find the x- and y-components after solving for the magnitude.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,814
5,594
V = ∫kdQ/r = -kQ/r
Eradial = -dV/dr
That doesn't work. In the integration you took r as constant, i.e. you really mean a in this context, not r.
The electric field is found by differentiating the voltage, as a function of position, with respect to position. You have effectively differentiated wrt changing radius of arc.
Try again.
 
  • #3
194
0
That doesn't work. In the integration you took r as constant, i.e. you really mean a in this context, not r.
The electric field is found by differentiating the voltage, as a function of position, with respect to position. You have effectively differentiated wrt changing radius of arc.
Try again.
So I was able to solve this by differentiating with respect to θ, and using dθ = ds/Q, and dq = 2Qds/πa. Then integrating dE = (kdq/a2)cos(Θ) for the x component, and replacing cos with sin for the y-component then substituted the values into dE.

I'm still slightly confused as to why my first approach did not work. I just finished the electric potential unit in my book and it recommends finding potential first if you have to solve for E because potential only requires scalar addition. Then I did an example where I had to find the potential from a ring of charge along the x-axis, where I used V = ∫kdQ/r = kQ/r and got the correct answer. How does finding the potential in this problem differ?
 
  • #4
194
0
That doesn't work. In the integration you took r as constant, i.e. you really mean a in this context, not r.
The electric field is found by differentiating the voltage, as a function of position, with respect to position. You have effectively differentiated wrt changing radius of arc.
Try again.
Or similarly, finding the potential at the center of the ring, you can use V = ∫kdQ/r = kQ/r.
 
  • #5
194
0
That doesn't work. In the integration you took r as constant, i.e. you really mean a in this context, not r.
The electric field is found by differentiating the voltage, as a function of position, with respect to position. You have effectively differentiated wrt changing radius of arc.
Try again.
Hmm. I might have misread your post. Is my potential correct, but my approach at finding electric field incorrect? If so, is there any way to find electric field from potential in this problem rather than the approach I took above?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,814
5,594
Hmm. I might have misread your post. Is my potential correct, but my approach at finding electric field incorrect? If so, is there any way to find electric field from potential in this problem rather than the approach I took above?
That's right, you have the potential correct but differentiating wrt the radius will not give you the field. To use the potential approach here you would need to find an expression for the potential as a function of position in relation to the origin, then differentiate wrt the position.
Easier here is to consider what direction the resultant field must be in, then, for each point on the arc (I assume it's an arc, not a quarter disc) find the component of the field it generates at the origin, in the desired direction. It looks like you have now done that, or something similar.
 

Related Threads on Electric field at the origin of a quarter-circle

  • Last Post
Replies
15
Views
2K
Replies
9
Views
20K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
8
Views
4K
Replies
1
Views
523
  • Last Post
Replies
3
Views
2K
Replies
6
Views
9K
Replies
2
Views
1K
Replies
10
Views
2K
Top