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# In a cube, given ${{E}_{x}}=5Ax+2B$, $A=10N/C.m$, $B=5N/C$ and $L=10cm$ .Find out: (A) Total electric flux through the cube(B) Net charge enclosed by the cube(C) What if the frame of reference is at the centre of the cube?

Last updated date: 01st Mar 2024
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Hint: We have been given the variation of the electric field along the x-axis. This means that only the faces along the x-axis are of our concern because the flux along all the other faces would be zero. Once we have found the total electric flux through the cube, we can use Gauss’ law to find the charge enclosed by the cube.

Formula Used:
$\phi =E.ds.\cos \theta$, $\phi =\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$

Complete step by step solution:
Since we know that electric flux passing through a surface due to electric field in the region is given as $\phi =E.ds.\cos \theta$ where $E$ is the electric field in the region, $ds$ is the area of cross-section of the surface and $\theta$ is the angle between the electric field and the surface through which the flux passes
Look at the figure below

Only the faces P and Q are along the x-axis, the remaining faces are perpendicular to the electric field and the cosine of $\text{90 }\!\!{}^\circ\!\!\text{ }$ is zero.
Now, to calculate the electric flux through the face P, we have the electric field ${{E}_{P}}=5Ax+2B$
Substituting the values, we get ${{E}_{P}}=5(10)(0)+2(5)=10N/C$ since at the face P $x=0$
Area of cross-section \begin{align} & (ds)={{L}^{2}}={{(10cm)}^{2}} \\ & \Rightarrow ds={{(0.1m)}^{2}}=0.01{{m}^{2}}\left[ \because 1cm=0.01m \right] \\ \end{align}
The electric flux through the face P,
\begin{align} & {{\phi }_{P}}=E.ds.\cos \theta \\ & \Rightarrow {{\phi }_{P}}=\left( 10N/C \right).\left( 0.01{{m}^{2}} \right).\cos 180{}^\circ \\ & \Rightarrow {{\phi }_{P}}=-0.1N.{{m}^{2}}/C\left[ \because \cos 180{}^\circ =-1 \right] \\ \end{align}
Similarly, the electric field at the face Q, ${{E}_{Q}}=5(10)(0.1)+2(5)=15N/C$ since at the face Q $x=L=0.1m$
Hence, electric flux through the face Q,
\begin{align} & {{\phi }_{Q}}=E.ds.\cos \theta \\ & \Rightarrow {{\phi }_{Q}}=\left( 15N/C \right).\left( 0.01{{m}^{2}} \right).\cos 0{}^\circ \\ & \Rightarrow {{\phi }_{Q}}=-0.15N.{{m}^{2}}/C\left[ \because \cos 0{}^\circ =1 \right] \\ \end{align}
Now, total electric flux through the cube
\begin{align} & \phi ={{\phi }_{P}}+{{\phi }_{Q}} \\ & \Rightarrow \phi =(0.15-0.01)=0.05N.{{m}^{2}}/C \\ \end{align}
Now, from Gauss’ law, we know that $\phi =\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$ where ${{q}_{in}}$ is the enclosed charge and ${{\varepsilon }_{0}}$ is the permittivity of free space
Hence ${{q}_{in}}=\phi \times {{\varepsilon }_{0}}$ where ${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}F/m$
Substituting the values, we get
\begin{align} & {{q}_{in}}=(0.05)\times (8.85\times {{10}^{-12}}) \\ & \Rightarrow {{q}_{in}}=4.42\times {{10}^{-13}}C \\ \end{align}

If the frame of reference is at the centre of the cube, the total flux passing through the cube would be zero (since both the faces P and Q would have equal distances from the reference and would cancel the electric flux) and hence the charge enclosed by the cube would be zero.

Note: The permittivity of free space is a physical constant used often in electromagnetism. It represents the capability of a vacuum to permit electric fields. Gauss's law has a close mathematical similarity with several laws in other areas of physics, such as Gauss's law for magnetism and Gauss's law for gravity. Any inverse-square law can be formulated in a way similar to Gauss's law.