Recent content by zapman345

You were right. The potential I had calculated should be $$ \begin{align*} V\left(\vec{r}\right) & =\frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{r^{2}+d^{2}-2dr\cos\left(\theta\right)}}-\frac{1}{\sqrt{R^{2}+\left(\frac{rd}{R}\right)^{2}-2dr\cos\left(\theta\right)}}\right]\\ \end{align*}$$...

I did but that wouldn't make the ##\theta## component vanish cause I find that $$E_{\theta} =...

I am required to find the direction of the electric field on the surface of a grounded conducting sphere in the proximity of a point charge ##+q##. The distance between the center of the sphere and the point charge is ##d## and using the method of images we find that the charge of the sphere is...

Thank you so much for all your help!

Okay so then it becomes dE_y=\frac{1}{4\pi\epsilon_{0}}\cdot\frac{1}{r^{2}}\sin\left(\theta\right)\lambda_{0}\sin^{2}\left(\varphi\right)Rd\varphi which then results in \begin{eqnarray*} E_{y} & = &...

Oh what I tried to say was that an element of the electric field in the y direction was equal to the y-component of the electric field vector. Sorry if I'm a bit slow to understand but the resulting electric field would be...

So it should be dE_y=\sin(φ)\cos(θ)dE \hat{y} then? Because that would be the y-component of a vector in spherical coordinates.

Now when I try to calculate the electric field in the y-direction I get E_{y} = \frac{\lambda_{0}}{4\pi\epsilon_{0}}\cdot\frac{R^{2}}{\left(z_{0}^{2}+R^{2}\right)^{\frac{3}{2}}}\int_{0}^{\pi}\sin\left(\varphi\right)d\varphi (following the same procedure as I did in the first post) my question...

Okay so an element at S produces an electric field in point P like E=-E_x \hat{x}-E_y \hat{y}+E_z \hat{z} and an element at S' produces E=+E_x \hat{x}-E_y \hat{y}-E_z \hat{z} thus adding both of those we have a total electric field of E=-2E_y \hat{y} ?

Because of the charge distribution there will be a negatively charged lower ring (φ=π to φ=2π) and a positively charged upper ring (φ=0 to φ=π). The negative semi-ring will have a negative z-component; a positive x-component and a negative y-component (?) and the positive semi-ring will have a...

Homework Statement Calculate the magnitude and direction of the electric field due to the ring in the point P on the z-axis. Homework Equations Charge distribution on the ring is given by λ(φ)=λ_0\cdot sin(φ). This should result in the total charge on the ring being zero. Electric field is...