Question
Question: What is the capacitance of the capacitor of square plates of area \(A\), shown in the figure , named after the English physicist Michael Faraday. A 1 farad capacitor, when charged with 1 coulomb of electrical charge, has a potential difference of 1 volt between its plates.
We know that the charge Q produced due to capacitance C and potential difference V is given as Q=CV. Also, the energy of the capacitor is E=21CV2.
We know that C=dKAϵ0 where, A is the area of the capacitance and d is the width of the dielectric Kand the electric constant ϵ0=8.854×10−12Fm−1
Here, we have a capacitanceC1 filled with K1, then, C1=dK1dϵ0×2l=2K1ϵ0l
Similarly, we have a capacitance C2 is filled with K2, then, we can divide the capacitance into two areas, then we have the following C2=2(2l)K2ϵ0ld+2dK2ϵ0ld
⟹C2=K2ϵ0d+2K2ϵ0ld
∴C2=K2ϵ0d(1+2l)
Then the net C=C1+C2=2K1ϵ0l+K2ϵ0d(1+2l)
Note:
The series of capacitors is the sum of reciprocal of its individual capacitors, whereas in resistance the parallel is the sum of reciprocal of its individual resistors. Also remember that capacitors can charge and discharge. To solve the question, we can divide the given area into small capacitance.