# Electric field at the origin of a quarter-circle

## 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.

V = ∫kdQ/r

## 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.

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
V = ∫kdQ/r = -kQ/r
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.

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?

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.

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?

haruspex