Question

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

Answer 1

Hint: Electric flux is the product of the electric field over the region and the area of the region. The electric field is obtained from the given relation. The product of the obtained electric flux and the permittivity of the free space is known as the charge.

Formula used:
(1) The formula of the electric flux is given by
$F = EA$
Where $F$ is the electric flux, $E$ is the electric field in a region and $A$ is the area of the region.
(2) The formula of the charge is given by
$q = { \in _0}F$
Where $q$ is the charge and ${ \in _0}$ is the permittivity of the free space.

Complete step by step answer:
Electric field, $\vec E = \dfrac{{{E_ \circ }x}}{l}\hat i$
Length of the field, $l = 2\,cm$
The constant, $a = 1\,cm$
${E_0} = 5 \times {10^3}\,N{C^{ - 1}}$
From the figure, it is clear that the electric flux mainly passes through the surface areas of the region.
Using the formula of the electric flux in a region,
$F = EA$
Substituting the known values in the above formula, we get
$F = \dfrac{{{E_ \circ }x}}{l} \times A$
$F = \dfrac{{5 \times {{10}^3}a}}{l} \times {a^2}$
By simplifying the above step, we get
$F = \dfrac{{5 \times {{10}^3}{{\left( {0.01} \right)}^{ - 3}}}}{{2 \times {{10}^{ - 2}}}}$
By performing the basic arithmetic operation, we get
$F = 2.5 \times {10^{ - 1}}$
By using the formula (2)
$q = { \in _0}F$
Substituting known values,
$q = 8.85 \times {10^{ - 12}} \times 2.5 \times {10^{ - 1}}$
$q = 2.2125 \times {10^{ - 12}}\,C$

Hence, the charge contained in the cubical volume is $2.21 \times {10^{ - 12}}\,C$.

Note: ${ \in _0}$ represents the permittivity of the free space and its value is taken as $8.85 \times {10^{ - 12}}$ . This value is substituted in the above calculation to obtain the result of the charge. The electric field is considered a vector since it contains both magnitude and direction.