A) \[\frac{{{Q}^{2}}}{2{{V}^{2}}}\]
B) \[\frac{1}{2}{{\varepsilon }_{0}}\frac{{{V}^{2}}}{{{d}^{2}}}\]
C) \[\frac{1}{2}\frac{{{V}^{2}}}{{{\varepsilon }_{0}}{{d}^{2}}}\]
D) \[\frac{1}{2}{{\varepsilon }_{0}}\frac{{{V}^{2}}}{{{d}^{2}}}\]
Correct Answer: D
Solution :
| [d] Key Idea: Energy stored between the plates of a capacitor is equal to \[\frac{1}{2}\frac{{{Q}^{2}}}{C}\]. |
| Energy stored, \[U=\frac{1}{2}\frac{{{Q}^{2}}}{C}\] |
| but \[\sigma =\frac{Q}{A}\] and \[C=\frac{{{\varepsilon }_{0}}A}{d}\] |
| \[\therefore \] \[U=\frac{1}{2}\frac{{{(\sigma A)}^{2}}}{({{\varepsilon }_{0}}A/d)}\] |
| or \[U=\frac{A{{\sigma }^{2}}d}{2{{\varepsilon }_{0}}}\] |
| or \[U=\frac{1}{2}{{\left( \frac{\sigma }{{{\varepsilon }_{0}}} \right)}^{2}}\times {{\varepsilon }_{0}}\,Ad\] |
| or \[U=\frac{1}{2}\,E_{{{\varepsilon }_{0}}}^{2}Ad\] |
| Energy stored per unit volume i.e., energy density in thus given by |
| \[u=\frac{U}{V}=\frac{U}{Ad}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\] |
| \[=\frac{1}{2}{{\varepsilon }_{0}}{{\left( \frac{V}{d} \right)}^{2}}=\frac{1}{2}\frac{{{\varepsilon }_{0}}{{V}^{2}}}{{{d}^{2}}}\] |
| Note: \[\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}\] is also a force on a conductor per unit area which is everywhere along the outward drawn normal to the surface. |
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