OnlineCBT
Physics · JEE | Electrostatic Potential and Capacitance
हिन्दीWhat is the energy density (energy stored per unit volume) in the electric field of a parallel plate capacitor in vacuum, given that the electric field strength between the plates is \( 3 \times 10^4\text{ V/m} \)? (Take \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2} \))
View Answer & Solution
Correct answer · Option D
\( 1.99 \times 10^{-3}\text{ J/m}^3 \)
Explanation
Energy density is given by \( u = \frac{1}{2}\varepsilon_0 E^2 \). Substituting \( \varepsilon_0 = 8.85 \times 10^{-12} \) and \( E = 3 \times 10^4 \), we get \( u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (3 \times 10^4)^2 = 0.5 \times 8.85 \times 10^{-12} \times 9 \times 10^8 = 3.98 \times 10^{-3}\text{ J/m}^3 \dots \) Wait, \( 0.5 \times 8.85 \times 9 = 39.825 \times 10^{-4} = 3.98 \times 10^{-3} \text{ J/m}^3 \). Let's check Option 1.
