Question

Calculate the electric flux through a cube of side ‘a’ as shown where $E_x=b{x}^{{\displaystyle \frac{1}{2}}}$ , $E_y=E_z=0$ , $a=10cm$ and $b=800NC-m^{{\displaystyle \frac{1}{2}}}$ .

Answer 1

Hint :Electric flux is only given in the x axis direction so we have to find only that side of flux. In order to solve this we need to take both the flux developed due to each side. For taking one side of length and another one for side length $2a$ . Putting the value of a and b will give the desired result of the flux of the cube. They are considered to originate from positive charges and have to terminate on negative charges. The area may be an arbitrary surface. Electric flux is a scalar quantity and electric fields are perpendicular to the surface area.

Complete Step By Step Answer:
In the given question since electric field is given only in the x axis, flux is equal to dot product of electric field and surface area.
So here, ${\varphi }_E=\overrightarrow{E}\bullet △\overrightarrow{S}$
Given; $E_x=b{x}^{{\displaystyle \frac{1}{2}}}$ and $a=10cm$
Now here,
 $$\begin{gathered} {\varphi }_L=\overrightarrow{E_L}\bullet △\overrightarrow{S} \\ =b{x}^{{\displaystyle \frac{1}{2}}}\times a^2{\cos }^2{180}^{\circ } \\ =-a^{2}b{x}^{{\displaystyle \frac{1}{2}}} \end{gathered}$$
Also if we want to know the new side length which is
 $$\begin{gathered} x=a+a \\ =2a \end{gathered}$$
Also for the second flux formed;
 $$\begin{gathered} {\varphi }_R=\overrightarrow{E_R}\bullet △\overrightarrow{S} \\ =b{x}^{{\displaystyle \frac{1}{2}}}\times a^2\cos 0^{\circ } \\ =-a^{2}b{x}^{{\displaystyle \frac{1}{2}}} \\ =-a^{2}b(2a{)}^{{\displaystyle \frac{1}{2}}} \end{gathered}$$
So net flux formed here in the cube,
 $$\begin{gathered} \varphi =-a^{2}b(a{)}^{{\displaystyle \frac{1}{2}}}-a^{2}b(2a{)}^{{\displaystyle \frac{1}{2}}} \\ =-a^{2}b\sqrt{a}(1+\sqrt{2}) \end{gathered}$$
Putting the value of a and b in the above equation we get;
 $$\begin{gathered} =-(10\times {10}^{-2})\times 800\times (10\times {10}^{-2}{)}^{{\displaystyle \frac{1}{2}}}\times 2.1414 \\ =6.107N{C}^{-1}{m}^{{\displaystyle \frac{1}{2}}} \end{gathered}$$

Note :
Basically electric flux is the property of an electric field in which electric field lines are inserted in a particular number of areas. They are considered to originate from positive charges and have to terminate on negative charges. The area may be an arbitrary surface. Electric flux is a scalar quantity and electric fields are perpendicular to the surface area.