
Prove that, electric potential energy per unit volume of a charged condenser is \[\dfrac{1}{2}{\varepsilon _0}{E^2}\] where the symbols have their usual meanings.
Answer
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Hint: In electromagnetism, when a capacitor is charged by a source of voltage, it starts to getting charged and at some point up to its capacity a capacitor is said to be fully charged when no more charge it can store and thus total energy stored in it is called electrostatic energy and we have to calculate this electrostatic energy per unit volume.
Complete step by step answer:
Let us suppose we have a capacitor of capacitance $C$ and the area of plates of capacitor is $A$ and distance between the two plates is $d$. Let $V$ be the potential difference applied across the capacitor and $E$ be the electric field inside the capacitor then we know that $V = Ed$.
Also, the relation between capacitance, area of plates and distance between the plates of a capacitor is given by,
$C = \dfrac{{{\varepsilon _0}A}}{d}$
Now,We also know that, the energy stored in a capacitor is calculated as $U = \dfrac{1}{2}C{V^2}$ putting the values of $V = Ed$ and $C = \dfrac{{{\varepsilon _0}A}}{d}$ in the energy equation $U = \dfrac{1}{2}C{V^2}$. We will get,
$U = \dfrac{1}{2}\dfrac{{{\varepsilon _0}A}}{d}{E^2}{d^2}$
$\Rightarrow \dfrac{U}{{Ad}} = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Since, $Ad$ is the total volume of the condenser so the term $\dfrac{U}{{Ad}}$ can be read as energy per unit volume lets it’s denoted by ${U_c}$ then we have,
$\therefore {U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Hence, the energy per unit volume of a condenser is ${U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$ ,proved.
Note: It should be remembered that, this ${U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$ is the energy per unit volume when capacitor is not filled by any dielectric material if it gets filled by any dielectric material having dielectric constant value of $K$ the energy will be ${U_c} = \dfrac{1}{2}{\varepsilon _0}\dfrac{{{E^2}}}{{{K^2}}}$ which shows that electrostatic energy per unit volume gets decreased when capacitor is filled with dielectric material.
Complete step by step answer:
Let us suppose we have a capacitor of capacitance $C$ and the area of plates of capacitor is $A$ and distance between the two plates is $d$. Let $V$ be the potential difference applied across the capacitor and $E$ be the electric field inside the capacitor then we know that $V = Ed$.
Also, the relation between capacitance, area of plates and distance between the plates of a capacitor is given by,
$C = \dfrac{{{\varepsilon _0}A}}{d}$
Now,We also know that, the energy stored in a capacitor is calculated as $U = \dfrac{1}{2}C{V^2}$ putting the values of $V = Ed$ and $C = \dfrac{{{\varepsilon _0}A}}{d}$ in the energy equation $U = \dfrac{1}{2}C{V^2}$. We will get,
$U = \dfrac{1}{2}\dfrac{{{\varepsilon _0}A}}{d}{E^2}{d^2}$
$\Rightarrow \dfrac{U}{{Ad}} = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Since, $Ad$ is the total volume of the condenser so the term $\dfrac{U}{{Ad}}$ can be read as energy per unit volume lets it’s denoted by ${U_c}$ then we have,
$\therefore {U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$
Hence, the energy per unit volume of a condenser is ${U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$ ,proved.
Note: It should be remembered that, this ${U_c} = \dfrac{1}{2}{\varepsilon _0}{E^2}$ is the energy per unit volume when capacitor is not filled by any dielectric material if it gets filled by any dielectric material having dielectric constant value of $K$ the energy will be ${U_c} = \dfrac{1}{2}{\varepsilon _0}\dfrac{{{E^2}}}{{{K^2}}}$ which shows that electrostatic energy per unit volume gets decreased when capacitor is filled with dielectric material.
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