Question

A charge Q is spread uniformly along the circumference of acircle of radius R. A point
particlewith charge q is placed at the center of this circle.The total force exerted on the
particle q can be calculated by Coulomb's law:
A) just use R for the distance D) result of the calculation iszero
B) just use 2R for the distance E) none of the above
C) just use 2πR for the distance

Answer 1

Answer:

D) result of the calculation is zero

Explanation:

Coulomb's Law is valid for only point-like particles. Since the ring is not a point-like, then we have to choose an infinitesimal portion (ds) of the ring, apply the Coulomb's Law to this portion and then integrate over the ring to find the total force.

The small portion (dq) will have the same charge density as the ring itself. Furthermore, the length of the infinitesimal portion is equal to the radius times the corresponding angle, dθ.

$\lambda = \frac{Q}{2\pi R} = \frac{dq}{Rd\theta}\\dq = \frac{Qd\theta}{2\pi}$

Therefore, the force between the charge at the center and the small portion is

$dF = \frac{1}{4\pi\epsilon_0}\frac{qdq}{R^2} = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}$

Since force is a vector, we have separate its x- and y-components,

$dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\cos(\theta)\\dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\sin(\theta)$

Now, we can integrate both of them over the ring.

$F_x = \int\limits^{2\pi}_0 dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\cos(\theta)d\theta = 0\\F_y = \int\limits^{2\pi}_0 dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\sin(\theta)d\theta = 0$

Since the integration from 0 to 2π for sine and cosine functions results as zero.

Therefore, the force on the charge at the center of a uniformly distributed ring is equal to zero.