Electric field at the origin of a quarter-circle

1. Apr 26, 2015

henry3369

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
V = ∫kdQ/r

3. 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.

2. Apr 26, 2015

haruspex

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. Apr 26, 2015

henry3369

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. Apr 26, 2015

henry3369

Or similarly, finding the potential at the center of the ring, you can use V = ∫kdQ/r = kQ/r.

5. Apr 26, 2015

henry3369

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. Apr 27, 2015

haruspex

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.