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प्रश्न
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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उत्तर १
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.
उत्तर २
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`
संबंधित प्रश्न
A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) Field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
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