Question

A uniform electric field $E = 3000\,V/m$exists within a certain region. What volume of space contains energy equal to $1.00 \times {10^{ - 7}}\,J?$ Express your answer in cubic meters and in liters.

Answer 1

Hint:Firstly get some idea about the uniform electric field. A uniform electric field is one in which the strength of the field is constant at all places. The field strength does not change in a homogeneous electric field, and the field lines tend to be parallel and equidistant. They're all evenly spaced.

Formula used:
Volume $V = \dfrac{{2{U_E}}}{{{\varepsilon _ \circ }{E^2}}}$
Where, ${U_E} = $Energy Density of electric field, $E = $Electric field and ${\varepsilon _ \circ } = 8.8542 \times {10^{ - 12}}\,C{N^{ - 1}}{m^2}$.

Complete step by step answer:
A potential difference between two charged or conducting plates kept at a fixed distance from each other is frequently used to create uniform fields. The plates are separated by an insulating medium or dielectric medium that is absolutely homogeneous (pure). At the centre of the plates, the electric field will be uniform. It may, however, vary or be non-uniform towards the plates' ends.

Let us know about the energy density of the electric field. The total quantity of energy in a system per unit volume is known as energy density. The number of calories per gramme of food, for example. Low energy density foods deliver less energy per gramme of food, allowing you to eat more of them because there are less calories.
${U_E} = \dfrac{1}{2}{\varepsilon _ \circ }{E^2}$

Given: Electric field $E = 3000\,V/m$
Energy Density of electric field ${U_E} = 1 \times {10^{ - 7}}\,J$
As we know: Volume $V = \dfrac{{2{U_E}}}{{{\varepsilon _ \circ }{E^2}}}$
$V = \dfrac{{2(1 \times {{10}^{ - 7}}\,J)}}{{(8.8542 \times {{10}^{ - 12}}){{(3000)}^2}}}$
$\Rightarrow V = 2.51 \times {10^{ - 3}}\,{m^3}$
Now we to do conversion of this above volume into liters:
$V = (2.51 \times {10^{ - 3}}) \times \dfrac{{1000L}}{{1\,{m^3}}}$
$\therefore V = 2.5\,L$

Hence, the volume of space is $2.51 \times {10^{ - 3}}\,{m^3}$ or $2.5\,L$.

Note: Let us know some important points regarding the electric field. There is no such thing as a negative electric field. Because it is a vector, it has both negative and positive axes. When an electron is negatively charged, it receives a force that opposes the field's direction.