FIGURE EX$24.17$shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) $2q/{ϵ}_0$, (b) $q/{ϵ}_0$, (c) $0\text{,}$and (d) $5q/{ϵ}_0$.

(a) The quantity of energy field that moves through such a solid object is referred to as the induced voltage. The light beam through a plate is proportional to the charge inside the surface, so according Gauss's rule, which is calculated $\left(24.18\right)$in the form

${\mathrm{\Phi }}_{\mathrm{e}}=\text{∮}\overrightarrow{\overrightarrow{E}}\cdot d\overrightarrow{A}=\frac{Q_{\text{in}}}{{ϵ}_o}$

The electric rate is driven by the charge from the inside of the closed surface, as depicted. Because every flux related to charges outside the closed surface is zero, we form a closed area around the number of charges the flux equals to obtain the fluxes.

To design a closed field, the zeta potential must be sketched in a same method. We draw a closed surface it around surface ions of intensity $2q$only for a flow of $2q/{ϵ}_{or}$, as seen below.

(b) We design a complete area around total costs of $q$for a flux of $q/{ϵ}_{or}$. This may transpire if the dual penalties $3q$$\text{and}-2q$are mixed, so we draw a flat sphere above it parameters.

(c) We design a complete area around total costs of $0$for a flux of $0$. This may transpire if the dual penalties $2q\text{and}$$-2q$are mixed, so we draw a flat sphere above it parameters.

(d) We design a complete area around total costs of $5q$for a flux of $5q/{ϵ}_o$This may transpire if the dual penalties $3q\text{and}2q$are mixed, so we draw a flat sphere above it parameters.