Q131 / 180 · COMEDK 2024 (Memory Based) · Physics
A $5.80 \mu \mathrm{~F}$ parallel plate air capacitor has a plate separation of 5 mm and is charged to a potential difference of 400 V . The energy density in the region between the plates is
- A$283 \times 10^{-2} \mathrm{~J} / \mathrm{m}^3$
- B$2.83 \mathrm{~J} / \mathrm{m}^3$
- Correct answerC$2.83 \times 10^{-4} \mathrm{~J} / \mathrm{m}^3$
- D$283 \times 10^6 \mathrm{~J} / \mathrm{m}^3$
Solution
The energy density in the region between the plates is given by
$\begin{aligned}<br/>u & =\frac{1}{2} \varepsilon_0 E^2 \\<br/>& =\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2 \quad\left(\because E=\frac{V}{d}\right) \\<br/>\Rightarrow \quad u & =\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(\frac{400}{5 \times 10^{-3}}\right)^2 \\<br/>& =283.2 \times 10^{-12} \times 10^6 \\<br/>& =283.2 \times 10^{-6} \\<br/>& =2.83 \times 10^{-4} \mathrm{~J} / \mathrm{m}^3<br/>\end{aligned}$
$\begin{aligned}<br/>u & =\frac{1}{2} \varepsilon_0 E^2 \\<br/>& =\frac{1}{2} \varepsilon_0\left(\frac{V}{d}\right)^2 \quad\left(\because E=\frac{V}{d}\right) \\<br/>\Rightarrow \quad u & =\frac{1}{2} \times 8.85 \times 10^{-12} \times\left(\frac{400}{5 \times 10^{-3}}\right)^2 \\<br/>& =283.2 \times 10^{-12} \times 10^6 \\<br/>& =283.2 \times 10^{-6} \\<br/>& =2.83 \times 10^{-4} \mathrm{~J} / \mathrm{m}^3<br/>\end{aligned}$
Hence, the correct answer is option: 'C'.