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Since there is no charge inside the sphere, the potential satisfys the Laplace's Equation $$ \nabla^2 V(r,\theta) = 0. $$ Due to the symmetry in the angle $\phi$, we can expand the potential in $r$ …

in the first look, I thought there may be a chance to use Gauss's theorem for there is a region near the point P, the electric field is in the $z$ direction. What if we make Gaussian surface, the tin- …

The electric field near the surface: The sheet charge density on the spherical surface $$ \sigma = \frac{Q}{4\pi r^2}. $$ This renders an electric field $$ E = \frac{\sigma}{\epsilon_o} = \frac{Q}{4\ …

For a potential, you may choose an arbitrary reference point $\vec r_0$ $$ V(\vec r) - V(\vec r_0) = -\int_{\vec r_0}^{\vec r} \vec F(\vec r') \cdot d\vec \ell'. $$ For a cylindrical mass source, the …

Because there is a $\delta(r-a)$ in the $\rho(\vec{r})$, therefore as long as the integral is concerned, these two expressions give same answer to the result of integration. They can be differentiate …

The first method is not correct by assuming a shperical symmetric geometry. For an element charge locates on the cylinder $\vec{r}' = (r', \phi', z')$. It contribution to the field point $\vec{r} = (r …