# Could you please solve and explain this question using Gauss' Law ?

## Apr 29, 2017

Gauss' Law $\int {\int}_{A} m a t h b f E \cdot d m a t h b f A = {Q}_{e n c l o s e d} / \epsilon$ simplifies where there is symmetry. Basically where the $m a t h b f E$ field always points at normal to surface, we get: $E = \frac{Q}{\epsilon A}$.

So we use symmetric Gaussian surfaces.

Sphere

Here, Gaussian surface is a concentric sphere just at the surface of the real sphere.

${E}_{s} = \frac{Q}{\epsilon \setminus 4 \pi {R}^{2}} = \frac{Q}{\epsilon \pi} \frac{1}{4 {R}^{2}}$

Cylinder

We use a concentric cylindrical Gaussian surface, but placed near the mid-point (length-wise) of the cylinder and with height $t$.

The real cylinder's surface charge density is:

$\sigma = \frac{Q}{2 \times \pi {\left(2 R\right)}^{2} + 2 \pi \left(2 R\right) H}$ and $H = 2 R$

$= \frac{Q}{16 \pi {R}^{2}}$

So our Gaussian surface encloses charge:

$2 \pi \left(2 R\right) t \times \frac{Q}{16 \pi {R}^{2}} = Q \frac{t}{4 R}$.

We assume no E field through end caps of our surface, it's all going through the curved surface.

${E}_{c} = \frac{Q \frac{t}{4 R}}{\epsilon 2 \pi \left(2 R\right) t} = \frac{Q}{\epsilon \pi} \frac{1}{16 {R}^{2}}$

Disc

Here we use another concentric cylinder of small radius $\rho$, but this time due to geometry we assume all the $m a t h b f E$ field comes out via its endcaps. Our Gaussian surface contains charge $\frac{2 \times \pi {\rho}^{2}}{2 \times \pi {\left(3 R\right)}^{2}} Q$ as it has total area normal to the $m a t h b f E$ field of $2 \pi {\rho}^{2}$, ie both the end caps.

${E}_{d} = \frac{\frac{{\rho}^{2}}{{\left(3 R\right)}^{2}} Q}{\epsilon 2 \pi {\rho}^{2}} = \frac{Q}{\epsilon \pi} \frac{1}{18 {R}^{2}}$

Looks like (A) to me