Question

What is the net electric flux through the cylinder of the figure(Figure 1) ? Assume that Q=200nC and q=4nC.

Answer 1

Concept and reason

The main concepts used to solve the problem are electric flux, Gauss’s law, and charge enclosed.

Initially, convert the charge from Nano coulombs to coulombs. Finally, use the gauss’s law to calculate the flux.

Fundamentals

Electric Flux: It is the measure of the flow of the electric field through a given area. Mathematically, it is expressed as,

$\Phi = \int {\vec E \cdot d\vec S}$

Here, $E$ is the electric field, $S$ is the surface area.

Gauss’s Law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ${\varepsilon _0}$ . Mathematically, the statement can be written as,

$\Phi = \frac{{{q_{{\rm{enclosed}}}}}}{{{\varepsilon _0}}}$

Here, is the net charge enclosed in the Gaussian surface ${q_{{\rm{enclosed}}}}$ and ${\varepsilon _0}$ is the permittivity of free space.

According, to Gauss’s law, the electric flux through the cylinder is as follows:

$\Phi = \frac{{{q_{{\rm{enclosed}}}}}}{{{\varepsilon _0}}}$

Here, is the net charge enclosed in the Gaussian surface ${q_{{\rm{enclosed}}}}$ and ${\varepsilon _0}$ is the permittivity of free space.

The permittivity of free space value is $8.85 \times {10^{ - 12}}\;{{\rm{C}}^2}{\rm{/}}\left( {{\rm{N}} \cdot {{\rm{m}}^2}} \right)$ .

Convert the units for the enclosed charge from nC to C.

$\begin{array}{c}\\{q_{{\rm{enclosed}}}} = 4.0\;{\rm{nC}}\\\\ = {\rm{4}}{\rm{.0}}\;{\rm{nC}}\left( {\frac{{1\;{\rm{C}}}}{{{{10}^9}\;{\rm{nC}}}}} \right)\\\\ = 4.0 \times {10^{ - 9}}\;{\rm{C}}\\\end{array}$

Substitute $4.0 \times {10^{ - 9}}\;{\rm{C}}$ for ${q_{{\rm{enclosed}}}}$ and $8.85 \times {10^{ - 12}}\;{{\rm{C}}^2}{\rm{/}}\left( {{\rm{N}} \cdot {{\rm{m}}^2}} \right)$ for ${\varepsilon _0}$ in the equation $\Phi = \frac{{{q_{{\rm{enclosed}}}}}}{{{\varepsilon _0}}}$ and solve for $\Phi$ .

$\begin{array}{c}\\\phi = \frac{{4.0 \times {{10}^{ - 9}}\;{\rm{C}}}}{{8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{\rm{ / N}} \cdot {{\rm{m}}^2}}}\\\\ = 452{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{ / C}}\\\end{array}$

Ans:

The electric flux through the cylinder was calculated to be $452{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{ / C}}$ .