Answer:

Explanation:
The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

Where A = Area of the plate
d = distance between the plates
dielectric constant
Case 1:
When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a
Therefore the capacitance of the capacitor becomes:
.......................(1)
Case 2:
Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:
.........................(2)
Equating (1) and (2)

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Answer:

Explanation:
We can assume this problem as two concentric spherical metals with opposite charges.
We have also to take into account the formulas for the electric field and the capacitance. Hence we have

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating
![dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]](https://tex.z-dn.net/?f=dV%3DEdr%5C%5C%5C%5CV%3D%5Cint_%7BR_1%7D%5E%7BR_2%7DEdr%3D-%5Cint_%7BR_1%7D%5E%7BR_2%7D%5Cfrac%7BkQ%7D%7Br%5E2%7Ddr%5C%5C%5C%5CV%3DkQ%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D)
Hence, the capacitance is
![C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B1%7D%7Bk%5B%5Cfrac%7B1%7D%7BR_2%7D-%5Cfrac%7B1%7D%7BR_1%7D%5D%7D)
but R1=a and R2=b

HOPE THIS HELPS!!