The energy density in the electric field of a parallel plate capacitor with charge 𝑄Q and capacitance 𝐶C is given by:a) 𝑄22𝐶2CQ 2 b) 𝑄2𝐶CQ 2 c) 𝑄24𝐶4CQ 2 d) 𝑄28𝐶8CQ 2
Question
The energy density in the electric field of a parallel plate capacitor with charge 𝑄Q and capacitance 𝐶C is given by:a) 𝑄22𝐶2CQ 2 b) 𝑄2𝐶CQ 2 c) 𝑄24𝐶4CQ 2 d) 𝑄28𝐶8CQ 2
Solution
The energy (U) stored in a capacitor is given by the formula U = Q²/2C.
The energy density (u) is defined as the energy per unit volume. For a parallel plate capacitor, the volume (V) can be expressed as the product of the area of one plate (A) and the separation between the plates (d). So, V = Ad.
The capacitance (C) of a parallel plate capacitor is given by the formula C = εA/d, where ε is the permittivity of the material between the plates.
We can substitute the expression for C from the above equation into the equation for U to get U = Q²d/2εA.
Then, the energy density (u) is U/V = (Q²d/2εA) / (Ad) = Q²/2εAd².
Therefore, the energy density in the electric field of a parallel plate capacitor with charge Q and capacitance C is Q²/2εAd².
None of the options a) 𝑄²/2𝐶², b) 𝑄²/𝐶, c) 𝑄²/4𝐶⁴, d) 𝑄²/8𝐶⁸ match this result. It seems there might be a mistake in the question or the provided options.
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