A) 0.01 J/m3
B) 0.01 J/m3
C) 1.0 J/m3
D) 10 J/m3
Correct Answer: B
Solution :
Key Idea: Energy density in the energy per unit volume. The energy per unit volume or the energy density is given, by \[U=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\] ?(1) Where \[{{\varepsilon }_{0}}\] is permittivity of free space and \[E\] is electric field. Also \[E=\frac{V}{d}\] ?(2) \[=\frac{\text{Potential difference}}{\text{Distance between the plates}}\] From \[Eqs.\] (1) and (2), we have \[U=\frac{1}{2}{{\varepsilon }_{0}}{{\left( \frac{V}{d} \right)}^{2}}\] Given, \[V=300\] Volt, \[d=2\,mm=2\times {{10}^{-3}}\,m,\] \[{{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N{{m}^{2}}\] \[\therefore \] \[U=\frac{1}{2}\times 8.85\times {{10}^{-12}}\times {{\left( \frac{300}{2\times {{10}^{-3}}} \right)}^{2}}\] \[U=0.1\,J/{{m}^{3}}\] Note: We can also say that if electric field \[\overset{\to }{\mathop{E}}\,\] exists in some space, then the space is a store of energy whose amount per unit volume is equal to energy density.You need to login to perform this action.
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