A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

<p>The energy density ($u$) between the plates of a capacitor is given by the formula:</p> <p>$u = \frac{1}{2} \varepsilon_0 E^2$</p> <p>where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:</p> <p>$E = \frac{V}{d}$</p> <p>Given:</p> <p><p>Capacitance ($C$) = 1 µF = $1 \times 10^{-6} \, \text{F}$</p></p> <p><p>Potential Difference ($V$) = 20 V</p></p> <p><p>Distance ($d$) = 1 µm = $1 \times 10^{-6} \, \text{m}$</p></p> <p>First, calculate the electric field $E$:</p> <p>$$ E = \frac{V}{d} = \frac{20 \, \text{V}}{1 \times 10^{-6} \, \text{m}} = 2 \times 10^7 \, \text{V/m} $$</p> <p>Now plug this into the formula for energy density:</p> <p>$$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2 \times 10^7 \, \text{V/m})^2 $$</p> <p>Calculate $u$:</p> <p>$u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}$</p> <p>$u = 2 \times 8.85 \times 10^2$</p> <p>$u = 1770 \, \text{J/m}^3$</p> <p>This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$. Therefore, the correct option is:</p> <p><strong>Option A</strong>: $1.8 \times 10^3 \, \text{J/m}^3$</p>