P20R2015 – About Electric Flux

The net charge enclosed by surface $\left(a\right)$ is $Q_a=-q$ and therefore, by Gauss’s Law, the electric flux through that surface is equal to

\boxed{{\phi }_E=-\frac{q}{{\varepsilon }_0}}

The net charge enclosed by surface $\left(b\right)$ is $Q_b=+3q-3q=0$ and therefore, by Gauss’s Law, the electric flux through that surface is equal to

\boxed{{\phi }_E=0}

The net charge enclosed through surface $\left(c\right)$ is $Q_c=0$ and therefore, by Gauss’s Law, the electric flux through that surface is equal to

\boxed{{\phi }_E=0}

The net charge enclosed through surface $\left(d\right)$ is $Q_a=-2q$ and therefore, by Gauss’s Law, the electric flux through that surface is equal to

\boxed{{\phi }_E=-\frac{2q}{{\varepsilon }_0}}

The goal of this problem is to emphasize the fact that only charge enclosed by the closed surface matters, not any charge lying outside of it (even though it is tempting to want to account for it). In practice, you will not compute the electric flux itself much as it doesn’t provide much information. You will only compute it as a means to an end to derive the magnitude of the electric field $E$ created by a charge distribution that has enough symmetry to make Gauss’s Law a very quick method to find the electric field.

The surface is closed and we therefore begin by drawing the (outward) normal vectors $d\overrightarrow{A_1}$ and $d\overrightarrow{A_2}$ for the circular base $A_1$ and the conical surface $A_2$ respectively.

Since the electric field lines remain parallel, every field line that enters through the base eventually exists through the conical side causing the net electric flux through the cone to be zero. This amounts to stating that the charge enclosed by the cone is zero: if it were not, then the fields lines would not remain parallel. Thus, by arguing that the net electric flux through the cone is zero, we derive the following

\oiint_{cone}{\overrightarrow{E}\cdot d\overrightarrow{A}}=\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A_1}}+\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A_2}}=0

The flux through the conical side $A_2$ can then be easily computed from the flux through the circular base $A_1=\pi R^2$. Indeed, we have

\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A_2}}=-\iint_{A_1}{\overrightarrow{E}\cdot d\overrightarrow{A_1}}=-\iint_{A_1}{E}{\mathrm{cos} \left(180\right)\ }dA_1=\iint_{A_1}{EdA_2}

The circular base has area $\pi R^2$ and the electric field is constant over the entire base and therefore the electric flux through the conical side is equal to

\boxed{\iint_{A_2}{\overrightarrow{E}\cdot d\overrightarrow{A_2}}=\iint_{A_1}{EdA_2}=E\pi R^2}

where the positive expression is consistent with outward flux.