Question

The electric field in a region is given by \[\overrightarrow E = \dfrac{{{E_o}x}}{l}\widehat i\]. Find the charge contained inside a cubical volume bounded by the surfaces $x = 0,x = a,y = 0,y = a,z = 0,z = a$. Take ${E_0} = 5 \times {10^3}N/C,l = 2cm$ and $a = 1cm$.

Answer 1

Hint: From the vector expression of electric field given, we will estimate the electric flux related to each surface. Then as you have guessed, we have to apply Gauss’s law to compute the charge contained inside the provided cubical volume.

Complete step by step solution:
Let’s start with electric flux. Total number of electric lines of force passing through a given surface normally is known as electric flux.
Electric flux through a small surface $dS$ for the electric field $\overrightarrow E $ is defined as the dot product between the vectors. Mathematically,
$d\phi = \overrightarrow E .\overrightarrow {dS} $
$ \Rightarrow d\phi = EdS\cos \theta $
Where $d\phi $ is the electric flux linked with the small surface $dS$.
$\theta $ is the angle between the surface vector and the electric lines of force.
Now electric flux through the whole surface will be,
$\phi = \int\limits_s {\overrightarrow E } .\overrightarrow {dS} $ ………. (1)
In the question the given electric field is \[\overrightarrow E = \dfrac{{{E_o}x}}{l}\widehat i\], which means that its direction is along the positive x-axis. Thus it is perpendicular to the surfaces $y = 0,y = a$ and $z = 0,z = a$.
So the electric flux linked with both of these surfaces will be zero, because the electric lines of forces and the electric field will have an angle between them $\theta = 90^\circ $, and $\cos 90^\circ = 0$.
For the remaining surface, $x = 0,x = a$, we will calculate the electric flux.
So at $x = 0$, magnitude of electric field is $\left| {\overrightarrow E } \right| = \dfrac{{{E_0} \times 0}}{l} = 0$ after substituting the respective values.
At $x = 0$, magnitude of electric field is $\left| {\overrightarrow E } \right| = \dfrac{{{E_0} \times a}}{l}$.
For $x = 0$ the electric flux linked with that surface would be
$\phi (x = 0) = EdS\cos 0^\circ $
Putting the value of magnitude of electric field we get,
$ \Rightarrow \phi (x = 0) = 0$ ………. (2)
Similarly for the surface $x = a$,
$\phi (x = a) = EdS\cos 0^\circ $
$ \Rightarrow \phi (x = a) = E \times {a^2} \times 1$
$ \Rightarrow \phi (x = a) = \dfrac{{{E_0}a}}{l} \times {a^2}$
$ \Rightarrow \phi (x = a) = \dfrac{{{E_0}{a^3}}}{l}$ ………. (3)
Now the total electric flux linked with the surfaces will be,
$\phi \left( {x = 0} \right) + \phi \left( {x = a} \right) = \dfrac{{{E_0}{a^3}}}{l}$
Gauss’s law states that total electric flux over a closed surface is $\dfrac{1}{{{\varepsilon _0}}}$ times the net charge enclosed by the closed surface, mathematically,
$\phi = \int\limits_s {\overrightarrow E .\overrightarrow {dS} } = \dfrac{q}{{{\varepsilon _0}}}$ ………. (4)
Where $q$ is the charge enclosed by the closed surface.
Now using the Gauss’s law, the net charge enclosed by the closed surface is
$q = \phi \times {\varepsilon _0}$
Substituting the values we get,
$ \Rightarrow q = \dfrac{{{E_0}{a^3}}}{l} \times {\varepsilon _0}$
Putting the values,
$ \Rightarrow q = \dfrac{{5 \times {{10}^3} \times {{(0.01)}^3}}}{{0.02}} \times 8.854 \times {10^{ - 12}} = 2.21 \times {10^{ - 12}}C$
So the net charge enclosed by the cubical surface is $2.21 \times {10^{ - 12}}C$.

Note: Notice the $\widehat i$ denotes that the electric field is along the x-axis, if it were $\widehat j$ it will be along y-axis and for $\widehat k$ it will be along z-axis. Also notice that the system of units when calculating the final answer we converted the system of units from c.g.s to S.I. also remember that Gauss’s law is applicable to closed surfaces only which contains an amount of charge.