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Question: What is the capacitance of the capacitor of square plates of area \(A\), shown in the figure ![](h...

What is the capacitance of the capacitor of square plates of area AA, shown in the figure

Explanation

Solution

A capacitor is a two terminal component that stores electrical energy in the form of potential energy, and later discharges them. This property is called the capacitance of the capacitor. The capacitors can be connected in series or in parallel with respect to each other.

Formula:
C=KAϵ0dC=\dfrac{KA\epsilon_{0}}{d}

Complete step by step answer:
A capacitor can store electrical energy, and behaves as a temporary battery. They are used mainly to maintain the power supply while batteries are being changed. It can also store information in the form of binary digits. It is the main component used in full wave and half wave rectifiers. (symbol: F), 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 QQ produced due to capacitance CC and potential difference VV is given as Q=CVQ=CV. Also, the energy of the capacitor is E=12CV2E=\dfrac{1}{2}CV^{2}.
We know that C=KAϵ0dC=\dfrac{KA\epsilon_{0}}{d} where, AA is the area of the capacitance and dd is the width of the dielectric KKand the electric constant ϵ0=8.854×1012Fm1\epsilon_{0}=8.854\times 10^{-12}Fm^{-1}
Here, we have a capacitanceC1C_{1} filled with K1K_{1}, then, C1=K1dϵ0d×l2=K1ϵ0l2C_{1}=\dfrac{K_{1}d\epsilon_{0}}{d}\times\dfrac{l}{2}=\dfrac{K_{1}\epsilon_{0}l}{2}
Similarly, we have a capacitance C2C_{2} is filled with K2K_{2}, then, we can divide the capacitance into two areas, then we have the following C2=K2ϵ0ld2(l2)+K2ϵ0ld2dC_{2}=\dfrac{K_{2}\epsilon_{0}ld}{2\left(\dfrac{l}{2}\right)}+\dfrac{K_{2}\epsilon_{0}ld}{2d}
    C2=K2ϵ0d+K2ϵ0ld2\implies C_{2}=K_{2}\epsilon_{0}d+\dfrac{K_{2}\epsilon_{0}ld}{2}
C2=K2ϵ0d(1+l2)\therefore C_{2}=K_{2}\epsilon_{0}d\left(1+\dfrac{l}{2}\right)
Then the net C=C1+C2=K1ϵ0l2+K2ϵ0d(1+l2)C=C_{1}+C_{2}=\dfrac{K_{1}\epsilon_{0}l}{2}+K_{2}\epsilon_{0}d\left(1+\dfrac{l}{2}\right)

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.