Question

Q1.A parallel plate capacitor is charged, such that electric field between the plates of capacitor is E. The energy density of the capacitor is​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Energy Density of a PARALLEL PLATE CAPACITOR:

Energy density can be defined as the total energy per unit volume of the parallel plate capacitor.

We know that energy in a parallel plate capacitor is as follows:

u=1/2CV²

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$\implies U = \dfrac{1}{2} \bigg( \dfrac{A \epsilon_{0} }{d} \bigg) V^{2}$

Multiplying and dividing by "d" :

$\implies U = \dfrac{1}{2} \bigg( \dfrac{A \epsilon_{0} \times d}{ {d}^{2} } \bigg) V^{2}$

$\implies U = \dfrac{1}{2} ( A \epsilon_{0} \times d){ \bigg( \dfrac{V}{d} \bigg) }^{2}$

$\implies U = \dfrac{ \epsilon_{0} }{2} ( A \times d){ \bigg( \dfrac{V}{d} \bigg) }^{2}$

Now, A × d = volume of capacitor

$\implies \dfrac{U}{Volume} = \dfrac{ \epsilon_{0} }{2} { \bigg( \dfrac{V}{d} \bigg) }^{2}</p><p>$

$\implies \dfrac{U}{Volume} = \dfrac{ \epsilon_{0} }{2} { \bigg( E\bigg) }^{2}</p><p>$

Let energy density be \muμ .

$\implies \mu= \dfrac{ \epsilon_{0} }{2} { \bigg( E\bigg) }^{2}$

So, final answer is :

$\boxed{ \bf \mu= \dfrac{ \epsilon_{0} }{2} { \bigg( E\bigg) }^{2} } </p><p>$