The capacitor is a parallel plate capacitor with a vacuum between the square plates. What is the length of a side if the distance between the plates is 0.250 mm

What is the capacitance of a parallel plate capacitor in vacuum? Sol. The basic definition for capacitance is: C = Since the electric potential can also be written as V = Ed Where, E is the electric field contribution of the parallel plate (two plane charge): = ol E0 %3D Sigma is the surface charge density given by: O= By, substitution, we get the capacitance of the parallel plate to be:

A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d. A parallel-plate capacitor is comprised of two square plates with sides of length ℓ and separation distance d. There are three dielectric materials in the space between the capacitor plates. A material with length ℓ⁄2, height d, and dielectric constant ?1 fills the left half of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?2 fills the top right quarter of the space. A material with length ℓ⁄2, height d⁄2, and dielectric constant ?3 fills the bottom right quarter of the space.   (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.)C =        (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.

Consider a parallel-plate capacitor with some non-zero charge on the plates.  If we double the charge surface density on each of the plates and double the separation between the plates, how does the energy density of the capacitor change?   A.  The energy density stays the same.   B.  The energy density is halved.   C.  The energy density decreases by a factor of 2–√.   D.  The energy density increases by a factor of 4.   E.  The energy density increases by a factor of 2–√.