MHT CET · Physics · Capacitance
The mean electrical energy density between plates of a charged air capacitor is (where \(\mathrm{q}=\) charge on capacitor, \(\mathrm{A}=\) Area of capacitor plate)
- A \(\frac{\mathrm{q}^2}{2 \varepsilon_0 \mathrm{~A}^2}\)
- B \(\frac{\mathrm{q}}{2 \varepsilon_0 \mathrm{~A}^2}\)
- C \(\frac{\mathrm{q}^2}{2 \varepsilon_0 \mathrm{~A}}\)
- D \(\frac{\varepsilon_0 A}{q^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{q}^2}{2 \varepsilon_0 \mathrm{~A}^2}\)
Step-by-step Solution
Detailed explanation
For a parallel plate capacitor, the energy density \(=\frac{1}{2} E^2 \varepsilon_0\)
But \(\mathrm{E}=\frac{\sigma}{\varepsilon_0}\)
\(\begin{aligned}
\therefore \text {Energy density } & =\frac{1}{2} \frac{\sigma^2}{\varepsilon_0^2} \times \varepsilon_0 \\
& =\frac{\sigma^2}{2 \varepsilon_0} \\
& =\frac{\left(\frac{q}{A}\right)^2}{2 \varepsilon_0} \quad \ldots(\because \sigma=q / A) \\
& =\frac{q^2}{2 A^2 \cdot \varepsilon_0}
\end{aligned}\)
But \(\mathrm{E}=\frac{\sigma}{\varepsilon_0}\)
\(\begin{aligned}
\therefore \text {Energy density } & =\frac{1}{2} \frac{\sigma^2}{\varepsilon_0^2} \times \varepsilon_0 \\
& =\frac{\sigma^2}{2 \varepsilon_0} \\
& =\frac{\left(\frac{q}{A}\right)^2}{2 \varepsilon_0} \quad \ldots(\because \sigma=q / A) \\
& =\frac{q^2}{2 A^2 \cdot \varepsilon_0}
\end{aligned}\)
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