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26. (a) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted between the plates How will.(i) it's capacitance, (ii) Electric field between plates, (ili). Charge (iv) energy stored change when battery remains connected ?
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26. (a) A parallel plate capacitor is charged by a battery to a potent...
Let the area of the plates be A,  and distance between them be d
Then capacitance before the insertion of the dielectric slab will be,
C = εoA/d
capacitance after the insertion of the dielectric slab will be,
C’ = εoKA/d
Let the battery charge the plate with charge Q. The electric field between the plates will be given by
Before the insertion of the dielectric slab
E = σ/2εo = Q/2εoA
After the insertion of the dielectric slab
E’ = σ/2εoK = Q/(2εoKA)
Energy stored in the capacitor is given by Initial energy before insertion of dielectric

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26. (a) A parallel plate capacitor is charged by a battery to a potent...
Capacitance:
When a dielectric slab is inserted between the plates of a parallel plate capacitor, the capacitance of the capacitor increases.

The capacitance of a parallel plate capacitor is given by the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

When a dielectric slab is inserted, it increases the capacitance because the presence of the dielectric material reduces the effective electric field between the plates. The polarization of the dielectric material creates additional charges on the surface of the dielectric facing the plates, which in turn creates an electric field that opposes the external electric field. This reduces the overall electric field between the plates and increases the capacitance.

Electric Field:
The electric field between the plates of a parallel plate capacitor decreases when a dielectric slab is inserted.

The electric field between the plates of a capacitor is given by the formula E = V/d, where V is the potential difference between the plates and d is the distance between the plates.

When a dielectric slab is inserted, the presence of the dielectric material reduces the electric field between the plates. This is because the polarization of the dielectric material creates additional charges on the surface of the dielectric facing the plates, which in turn creates an electric field that opposes the external electric field. The net effect is a decrease in the electric field between the plates.

Charge:
The charge on the plates of a parallel plate capacitor remains the same when a dielectric slab is inserted.

The charge on the plates of a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference between the plates.

When a dielectric slab is inserted, the capacitance of the capacitor increases, but the potential difference across the plates remains the same. As a result, the charge on the plates remains unchanged.

Energy Stored:
The energy stored in a parallel plate capacitor increases when a dielectric slab is inserted.

The energy stored in a capacitor is given by the formula U = 1/2 CV², where U is the energy stored, C is the capacitance, and V is the potential difference between the plates.

When a dielectric slab is inserted, the capacitance of the capacitor increases. Since the potential difference across the plates remains the same, the energy stored in the capacitor increases. This is because the increase in capacitance allows the capacitor to store more charge at the same potential difference, resulting in an increase in the energy stored.

Overall, when a dielectric slab is inserted between the plates of a parallel plate capacitor while the battery remains connected, the capacitance and energy stored increase, the electric field between the plates decreases, and the charge on the plates remains the same.
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26. (a) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted between the plates How will.(i) it's capacitance, (ii) Electric field between plates, (ili). Charge (iv) energy stored change when battery remains connected ?
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26. (a) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted between the plates How will.(i) it's capacitance, (ii) Electric field between plates, (ili). Charge (iv) energy stored change when battery remains connected ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 26. (a) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted between the plates How will.(i) it's capacitance, (ii) Electric field between plates, (ili). Charge (iv) energy stored change when battery remains connected ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 26. (a) A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted between the plates How will.(i) it's capacitance, (ii) Electric field between plates, (ili). Charge (iv) energy stored change when battery remains connected ?.
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