Forces from a cherged ring onb a charged particle

[hadoque]

Homework Statement

A charge, Q, is equally distributed in a small wireformed as a circle with radius R. Another charge, q, is placed at a distance R above the midpoint of the circle.
What force affects the charge q?

Homework Equations

$$F = \frac{Qq}{4 \pi \varepsilon_0 r^2}$$

The Attempt at a Solution

A distance , a, between a segment of the ring, dl, and q is $$a=\sqrt{2}R$$. We set the z-axis as the normal of the midpoint of the circle. We split the forces (if we think of Q as split into small segments, dQ) on q into forces along the z-axis and forces parallell to the xy-plane. The forces parallell to the xy-plane cancel each other out, and the forces along the z-axis add upp. Since $$a=\sqrt{2}R$$ the z-resultant of the force would have a factor $$2$$, the force on q add upp to $$F = \frac{Qq}{4 \pi \varepsilon_0 (R/\sqrt{2})^2}$$.

However, the answer in my problem collection says the answer should be $$F = \frac{Qq}{4 \pi \varepsilon_0 \sqrt{2}(\sqrt{2}R)^2}$$. Which answer is the right one?

 

[Jhero]

Dear forum post author,

Thank you for your question. After reviewing your solution attempt and the given answer in your problem collection, it appears that the given answer is correct. Let me explain why.

Firstly, your approach of splitting the forces into components along the z-axis and the xy-plane is correct. However, there is a slight error in your calculation of the z-resultant force. The distance, a, between a segment of the ring and q is not equal to R/sqrt(2), but rather it is equal to (R/sqrt(2))^2 + (R/sqrt(2))^2 = 2R^2. This is because we are dealing with a right triangle, and using the Pythagorean theorem, we can find the hypotenuse, which is the distance, a.

Therefore, the z-resultant force would be F = (Qq)/(4πε0a^2) = (Qq)/(4πε0(2R^2)). This is equivalent to the given answer, F = (Qq)/(4πε0√2(√2R)^2). The only difference is that the given answer has the distance, a, written in a simplified form.

I hope this clears up any confusion and helps you better understand the problem. Keep up the good work!