let a parallel plate capacitor be in the vacuum with one plate cathode located at x=0 emitting electrons which are collected at the other plate anode located at x=d the arrangement makes up a vacuum diode lets the volume charge density within the capacitor be given by pv(x)=-(4/9) E0V0d^-4/3 x^-2/3 C/m^3 where the cathode is located at x=0 and the anode at x=d . The voltage between the anode and the cathode is V0 and the total charge in 1m^2 cross section is -12 nC .1 find the Vo if d = 6mm . 2 find d if vo=5v .

SProblem 22: An insulating material with a dielectric constant of 4.37 and a dielectric strength of 6.3 MV/m will be used to fill the gap between the plates of a 6.25 nC parallel-plate capacitor. What is the minimum plate area, in squared meters, if the capacitor must withstand applied voltages up to 4.1 kV/m? Amin = m ,2

Two parallel plates of a capacitor are 4.0mmapartin air. The charge density on the capacitorplatesis 5.6 x 10-5C/m2. Calculate the energy density in between the plates.

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.