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Questions
Obtain the expression for capacitance for a parallel plate capacitor.
Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.
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Solution 1
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = `sigma/ε_0` where σ is the surface charge density on the plates σ = `"Q"/"A"`. If the separation distance d is very much smaller than the size of the plate (d2 << A), then the above result is used even for finite-sized
parallel plate capacitor.
Capacitance of a parallel plate capacitor
Capacitance of a parallel plate capacitor
The electric field between the plates is
E = `"Q"/("A"ε_0)` ....(1)
Since the electric field is unifonn, the electric potential between the plates having separation d is given by
V = Ed = `"Qd"/("A"ε_0)` ....(2)
Therefore the capacitance of the capacitor is given by
C = `"Q"/"V" = "Q"/(("Qd"/("A"ε_0))) = (ε_0"A")/"d"` ....(3)
From equation (3), it is evident that capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates. This can be understood from the following.
- If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
- If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.
Solution 2
Derivation of the expression for the capacitance.
Let the two plates be kept parallel to each other separated by a distance d and the cross-sectional area of each plate is A. Electric field by a single thin plate E = `sigma/(2epsi_0)`
The total electric field between the plates E = `sigma/epsi_0` = `Q/(Aepsi_0)`
The potential difference between the plates V = Ed = `[Q/(Aε_0)]`d.
Capacitance C = `"Q"/"V" = (Aepsi_0)/d`
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