The space between the plates of a parallel-plate capacitor, shown below, is filled with two slabs of different dielectric materials
Problem 5
The space between the plates of a parallel-plate capacitor, shown below, is filled with two slabs of different dielectric materials. The slab at the top has thickness 2d and a relative dielectric constant of er1 = 3 and the one at the bottom has thickness d and a relative dielectric constant of er2 = 2. The capacitor plates have surface area S.
a. Assume a total charge of +Q on the top plate and -Q on the bottom plate. Find electric flux density D in each slab using the boundary condition on the normal component of D.
b. Find the electric field E in each slab
c. Find the polarization P in each slab
d. Find the potential difference between the plates in terms of Q, d, e0
e. Find the surface charge density on both plates in terms of V, d, e0
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The space between the plates of a parallel-plate capacitor, shown below, is filled with two slabs of different dielectric materials
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Problem 7 The space between a parallel plate capacitor is filled with two slabs of dielectric material, as shown in figure (18.46) The dielectric constant of one slab is κι and the dielectric constant of the other slab is K2. The separation between the plates is d, and each slab fills half of the space between the plates of the capacitor. Determine the capacitance of this capacitor if the area of the two plates is A 2 Figure 18.46: Problem7
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Please answer the question on this image regarding a
parallel-plate capacitor that is half-filled with
dielectric
Problem 4 (6 points) A parallel-plate capacitor is half-filled with a slab of material of dielectric constant e, as shown in the figure. d/2 a) By what factor is the capacitance increased as a result of the presence of the dielectric? b) If the top and bottom conducting plates of the capacitor have surface charge density to and –o, respectively, identify the location, sign,...
7. A parallel plate capacitor has the space between plates filled with two slabs of dielectric (same size) with constants ki and ką. Show that the capacitances are C = 289A K; $2 for the left one and C d *y+ C = 91 Ni+2 for the right one, where d is the total spacing between plates.
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