Question

The electric field in a region is given by\[\overrightarrow E = \dfrac{{{E_0}x}}{l}\hat i\]. If the charge contained inside a cubical volume bounded by the surfaces \[x = 0,x = a,y = 0,y = a,z = 0,z = a\] is \[x \times {10^{ - 12}}C\]. Find x. (integer value) Take \[{E_0} = 5 \times {10^3}N/C\],l=2cm,a=1cm

Answer 1

Hint:
- Apply Gauss law.
- You should know the definition of Electric flux.
- You should know the fundamental numerical values.

Complete step by step solution:
Electric field is the region in which that force is felt.
 The electric field strength \[ = \] force per unit charge units \[ = \] newton’s per coulomb.
It is also defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. And Electric flux, property of an electric field that may be thought of as the number of electric lines of force \[\left( {or{\text{ }}electric{\text{ }}field{\text{ }}lines} \right)\] that intersect a given area.

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
The electric flux will contribute only due to surfaces \[x = 0\;and\;x = a.\] Because the electric field is perpendicular to other surfaces.
Thus,
\[
  \phi \left( {x = 0} \right) = \dfrac{{{E_0}\left( 0 \right)}}{l} \cdot {a^2} = 0 \\
  \phi \left( {x = a} \right) = \dfrac{{{E_0}\left( a \right)}}{l} \cdot {a^2} = \dfrac{{{E_0}}}{l} \cdot {a^3} \\
 \]
Using Gauss's law,
 \[
  \dfrac{q}{{{\varepsilon _0}}} = \phi \left( {x = a} \right) = \dfrac{{{E_0}}}{l} \cdot {a^3} \\
   \Rightarrow q = \dfrac{{{E_0}}}{l} \cdot {a^3}{\varepsilon _0} \\
  q = \dfrac{{5 \times {{10}^3}}}{{2 \times {{10}^{ - 2}}}} \cdot {10^{ - 6}} \times 8.854 \times {10^{12}} \\
  q = 2.21 \times {10^{ - 12}} \\
  q \sim 2 \times {10^{ - 12}}C \\
\]

Hence, the value of x is \[{\mathbf{2}}.{\mathbf{21}}\] (\[ \sim 2\])

Note:
- Since the charge is inside the cube so you have to take gauss’s law.
- Try to apply laws in these kinds of problems.