Question

The total electric flux from a cubical box 34.0 cm on a side is 1.36 103 N·m2/C. What charge is enclosed by the box?

Answer 1

Concepts and reason

The concepts required to solve this is Gauss law for electrostatics.

Initially, write the expression of the Gauss law. Rearrange the expression of the gauss law for the charge enclosed. Finally, substitute the values of the electric flux and the value of the permittivity, and calculate for charge enclosed by the box.

Fundamentals

The electric flux is defined as the number of electric field lines passing through a surface.

The expression for electric flux is,

${\Phi _E} = EA\cos \theta$

Here, E is the electric field, A is the surface area, and $\theta$ is the angle between the electric field and the normal to the plane.

The expression for Gauss’s law is,

${\Phi _E} = \frac{{{Q_{{\rm{enc}}}}}}{{{\varepsilon _0}}}$

Here, ${Q_{{\rm{enc}}}}$ is the total charge enclosed in the closed surface and ${\varepsilon _0}$ is the permittivity due to free space.

The expression of the electric flux is,

${\Phi _E} = \frac{{{Q_{{\rm{enc}}}}}}{{{\varepsilon _0}}}$

Here, ${Q_{{\rm{enc}}}}$ is the total charge enclosed in the closed surface and ${\varepsilon _0}$ is the permittivity due to free space

Rearrange the expression of for the charge enclosed.

${Q_{{\rm{enc}}}} = {\varepsilon _0}{\Phi _E}$

The expression for the charge enclosed is,

${Q_{{\rm{enc}}}} = {\varepsilon _0}{\Phi _E}$

Substitute $8.85 \times {10^{ - 12}}{\rm{ F}} \cdot {{\rm{m}}^{ - 1}}$ for ${\varepsilon _0}$and $1.36 \times {10^3}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}$ for ${\Phi _E}$.

$\begin{array}{c}\\{Q_{{\rm{enc}}}} = {\varepsilon _0}{\Phi _E}\\\\ = \left( {8.85 \times {{10}^{ - 12}}{\rm{ F}} \cdot {{\rm{m}}^{ - 1}}} \right)\left( {1.36 \times {{10}^3}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/C}}} \right)\\\\ = \left( {12.04 \times {{10}^{ - 9}}{\rm{ C}}} \right)\left( {\frac{{{{10}^9}{\rm{ nC}}}}{{1{\rm{ C}}}}} \right)\\\\{\rm{ = 12}}{\rm{.04 nC}}\\\end{array}$

Ans:

The charge enclosed by the box is ${\rm{12}}{\rm{.04 nC}}$.