baker265 said:Homework Statement
A parallel-plate capacitor with C = 16 mu or micro FF is charged so as to contain 1.7 J of energy. If the capacitor has a vacuum between plates that are spaced by 0.35 mm, what is the energy density (the energy per unit volume)?
I am a little lost of this problem and have no idea if I'm doing it correctly. Am I ?
Homework Equations
v = A*d
E,density=U/v
The Attempt at a Solution
v=16e-3/.35e-3=45.71 ?
E=1.7/45.71=.0371 J/m^3
The energy density of a capacitor refers to the amount of energy that can be stored in the capacitor per unit volume or unit mass. It is measured in joules per cubic meter (J/m3) or joules per kilogram (J/kg).
The energy density of a capacitor can be calculated by multiplying the capacitance (in farads) by the square of the voltage (in volts) and dividing by 2. This can be represented by the formula E = 1/2 * C * V^2.
The energy density of a capacitor is affected by several factors, including the type of dielectric material used, the distance between the plates, and the voltage applied. Generally, a higher dielectric constant, thinner distance between plates, and higher voltage can result in a higher energy density.
In general, capacitors have a lower energy density compared to other energy storage devices, such as batteries. This is because the energy density of a capacitor is dependent on the dielectric constant of the material, while batteries can store energy chemically, resulting in a higher energy density.
The energy density of a capacitor is important in various electronic devices, such as mobile phones, laptops, and cameras, where small and lightweight energy storage is needed. It is also used in electric vehicles, where high energy density capacitors are used for regenerative braking to store energy and improve efficiency.