Question

A point charge of 1.8μC at the center of a cubical Gaussian surface 55 cm on edge. What is the net electric flux through the surface?

Answer 1

Concepts and reason

The concept of Gauss’s law is used to solve this problem.

Initially, convert the unit of charge from ${\rm{\mu C}}$ to C by using the unit conversion. Finally, calculate the net electric flux through the surface by using gauss law.

Fundamentals

According to the Gauss’s law, the total electric flux in a closed surface is equal to the charge enclosed divided by the permittivity.

The Gauss’s law is,

$\phi = \frac{q}{{{\varepsilon _0}}}$

Here, $\phi$ is the electric flux, q is the charge, and ${\varepsilon _0}$is the permittivity of free space.

The charge enclosed in the center of the cubical gaussian surface is,

$\begin{array}{c}\\1.8\,{\rm{\mu C}} = 1.8\,{\rm{\mu C}}\left( {\frac{{{{10}^{ - 6}}\;{\rm{C}}}}{{1\;{\rm{\mu C}}}}} \right)\\\\ = 1.8 \times {10^{ - 6}}\;{\rm{C}}\\\end{array}$

The Gauss’s law is,

$\phi = \frac{q}{{{\varepsilon _0}}}$

Substitute $1.8 \times {10^{ - 6}}\;{\rm{C}}$ for q and $8.85 \times {10^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}$ for ${\varepsilon _0}$ in the above formula.

$\begin{array}{c}\\\phi = \frac{{1.8 \times {{10}^{ - 6}}\;{\rm{C}}}}{{8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^{\rm{2}}}{\rm{/N}} \cdot {{\rm{m}}^{\rm{2}}}}}\\\\ = 0.203 \times {10^6}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}/{\rm{C}}\\\end{array}$

The total flux in the Gaussian surface is $0.203 \times {10^6}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}/{\rm{C}}$.

Ans:

The total flux in the Gaussian surface is $0.203 \times {10^6}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}/{\rm{C}}$.