Question

A cube of marble having each side 1cm is kept in an electric field of intensity 300V/m. Determine the energy contained in the cube of dielectric constant 8. [Given: ${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$].

Answer 1

Hint: Consider the cubical dielectric material as a parallel plate capacitor. Then the energy stored is $E=\dfrac{1}{2}C{{V}^{2}}$. Use V=Ed to find V. Use the formula for capacitance i.e. $C=k\dfrac{{{\varepsilon }_{0}}A}{d}$ and find the capacitance. Finally, substitute the values of V and C to find E.

Formula used:
$E=\dfrac{1}{2}C{{V}^{2}}$
$C=k\dfrac{{{\varepsilon }_{0}}A}{d}$
V=Ed
$u=\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}$

Complete step-by-step answer:
It is given that a cube of marble is placed in an external electric field.
When a dielectric material is placed in an external electric field, the electric affects the atoms or charges inside the dielectric material. As a result, positive charges are deposited on one side and negative charges of equal magnitude are deposited on the other side. If the dielectric material is in the shape of a cube, then it acts as a parallel plate capacitor with a dielectric inserted in between the plates.
Let the potential difference across the parallel and opposite sides be V. The energy stored in the cubicle material is equal to $E=\dfrac{1}{2}C{{V}^{2}}$ ….. (i), where C is the capacitance of the cube (capacitor).
And the capacitance $C=k\dfrac{{{\varepsilon }_{0}}A}{d}$ …. (ii).
Here, k is the dielectric constant of the dielectric material, d is the distance between the opposite sides, A is the area of one of the opposites sides and ${{\varepsilon }_{0}}$ is the absolute permittivity of free space.
In the case of a cube, d is equal to the length of the side and A is equal to the area of one of the faces of the cube.
Therefore, d=1cm=0.01m and A=$1c{{m}^{2}}={{10}^{-4}}{{m}^{2}}$. The given value of k is 8.
Substitute the values of d, A and k in equation (ii).
 $C=8\times \dfrac{8.85\times {{10}^{-12}}\times {{10}^{-4}}}{0.01}=70.8\times {{10}^{-14}}F$
When the electric field (E) is constant, the potential difference (V) across a length of d is given as V=Ed.
In this case, $E=300V{{m}^{-1}}$ and d=0.01m
$\Rightarrow V=Ed=300\times 0.01=3V$.
Substitute the values of C and V in equation (i).
$E=\dfrac{1}{2}\left( 70.8\times {{10}^{-14}} \right){{(3)}^{2}}=318.6\times {{10}^{-14}}J$
Hence, the energy stored in the cube of marble is $318.6\times {{10}^{-14}}J$.

Note: There is an alternative way to solve the given question.
Energy density inside a material is given as $u=\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}$ …..(1), where E is the external electric field.
Energy density is the energy per unit volume of the material.
Let the total energy stored be U and the volume of the cube be V. Then $u=\dfrac{U}{V}$.
$\Rightarrow U=uV$ …. (2).
Substitute the values of E and k in equation (1).
$\Rightarrow u=\dfrac{1}{2}\times 8.85\times {{10}^{-12}}\times 8\times {{\left( 300 \right)}^{2}}=3.185\times {{10}^{-6}}J{{m}^{-3}}$.
Volume of the cube will be $V={{(0.01)}^{3}}{{m}^{3}}={{10}^{-6}}{{m}^{3}}$.
Substitute the value of u and V in equation (2).
$\Rightarrow U=3.185\times {{10}^{-6}}\times {{10}^{-6}}=3.185\times {{10}^{-12}}=318.5\times {{10}^{-14}}J$