- #1

- 373

- 94

I start with Coulomb's Law.

$$ dE(r) = \frac{1}{4\pi\epsilon}\frac{dq}{r^2} = \frac{\lambda R}{4\pi\epsilon}\frac{d\theta}{r^2},$$

in which ##\vec R## is a vector with magnitude equal to the radius of the disk that points from the origin to an infinitesimal charge ##dq##, and ##d\theta## is an angle of infinitesimal displacement that the vector ##\vec R## makes relative to the horizontal.

Then, I use the Law of Cosines to define ##r^2## in terms of ##\theta##,

$$ r^2 = R^2 + P^2 - 2RPcos(\theta - \phi), $$

in which ##\vec P## is a position vector that points from the origin to the point of interest with magnitude ##P=\sqrt{x^2 + y^2}##, ##\theta## is the angle ##\vec R## makes with the horizontal, ##\phi## is the angle ##\vec P## makes with the horizontal and ##\vec r## is the distance from the infinitesimal charge ##dq## to the point of of interest, so that ## \vec P - \vec R = \vec r ##.

Using the identity ## cos(\theta - \phi) = cos \theta cos \phi + sin \theta sin \phi ## and ##tan\phi = \frac{y}{x}##,

$$ r^2 = R^2 + P^2 - 2R(xcos\theta + ycos\theta),$$

$$ dE(P) = \frac{\lambda R}{4\pi\epsilon}\frac{d\theta}{R^2 + P^2 - 2R(xcos\theta + ycos\theta)},$$

$$ dE_x(P) = dE(P)cos\phi =\frac{\lambda R}{4\pi\epsilon} \frac{x}{P} \frac{d\theta}{R^2 + P^2 - 2R(xcos\theta + ycos\theta)} $$

and

$$ E_x(P) = dEcos\phi =\frac{\lambda R}{4\pi\epsilon} \frac{x}{P} \int \frac{d\theta}{R^2 + P^2 - 2R(xcos\theta + ycos\theta)} = 0$$ from ##\theta = 0 ## to ##2\pi##.

How is it possible?