Question

A closed surface encloses a net charge of $-3.6\mu C$. What is the electric flux through the surface?

Answer 1

Hint: A Gaussian surface is said to be an imaginary surface which is said to be considered in order to apply the formula for Gauss’ law. The concept of electric flux needs to be applied to determine the answer to the problem. The formula for the electric flux in accordance with Gauss’ law for electric charges is applied in order to find the electric flux through the surface.

Formula used:
The formula for the electric flux given by Gauss’ law is:
${\varphi }_{net}=q_{enclosed}\times {\epsilon }_0$
Where, $q_{enclosed}$ is the charge that is enclosed and ${\epsilon }_0$ is the permittivity of free space.

Complete step by step answer:
 The above problem revolves around the concept of electric flux and Gauss' law. In order to find out the electric flux through the surface we first need to know the concept behind electric flux and Gauss' law.Where there is a charge said to be placed in the region of an electric field then there is some force that is experienced by it which was given by the Coulomb’s law.

The measurement or the amount of electric field that is present due to which the charge experiences some force is given by a quantity known as electric lines of force. Thus, the concept of electric flux comes into picture. Electric flux for a given area is defined as the measure of the total electric lines of force passing normally through that area inside an electric field. The term flux means flow and this flux is a property for any vector field. Electric flux is said to be the property of electric fields.In order to calculate this net flux across any given area or surface, Gauss' law was introduced.

The Gauss’ law gives the relationship between the total flux passing through any closed surface and the net charge that is enclosed. The Gauss’ theorem states that the total flux through a closed surface is ${\displaystyle \frac{1}{{\epsilon }_0}}$ times the net charge enclosed within the surface. This theorem is valid only for charges that are enclosed by the shape of the closed surface.

Since, in order to apply Gauss’ law we require a closed surface and this is where the concept of a Gaussian surface comes into picture. An imaginary shape that is drawn around the surface in which the total flux needs to be found is called the Gaussian surface. This can be drawn in any shape or size and the charge that is enclosed inside this surface is what is considered.

The formula for Gauss’ law for the charge enclosed inside a surface is given by the equation:
${\varphi }_{net}={\displaystyle \frac{q_{enclosed}}{{\epsilon }_0}}$ ----------($1$)
Here, $q_{enclosed}$ indicates the total or the sum of all the charges that are present inside the closed surface that is considered.

Let us now extract the data given in the question. The question specifies the total charge that is enclosed. The permittivity of space is said to be a constant with a particular fixed value. We are asked to find the total electric flux passing through the closed surface.

Given, $q_{enclosed}=-3.6\mu C$ and ${\epsilon }_0=8.855\times {10}^{-12}$.We now convert the charge in micro- coulomb to Coulomb as it is the SI unit of charge. We know that,
$1\mu C={10}^{-6}C$
Thus, $-3.6\mu C=-3.6\times {10}^{-6}C$
We now substitute these given values in the Gauss’ law equation. Hence, by substituting the above values in equation ($1$) we get:
${\varphi }_{net}={\displaystyle \frac{-3.6\times {10}^{-6}}{8.855\times {10}^{-12}}}$
By simplifying further we get:
${\varphi }_{net}=-0.4065\times {10}^{12-6}$
$\Rightarrow {\varphi }_{net}=-4.065\times {10}^{5}N{m}^2{C}^{-1}$
We now approximate or round off this answer to $3$ significant figures and hence we get:
$\Rightarrow {\varphi }_{net}\approx -4.07\times {10}^{5}N{m}^2{C}^{-1}$
Hence, this is the total electric flux through the closed surface.

Note: The charge that is considered should be strictly present inside the Gaussian surface that has been considered and this is the criteria for which the Gauss’ law is applied.The charges outside the Gaussian surface are not taken into consideration. Usually, the Gaussian surface is taken to be a regular shape instead of an irregular one to make calculation easier.