A parallel plate capacitor with air as dielectric is connected to a power supply and charged to a potential difference Vo. After disconnecting from power supply, a sheet of insulating material is inserted between the plates completely filling the space between them. How will its
(i) Capacity (ii) electric field and (iii) energy change.
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The charge on plates remains same.
capacitance becomes K times. K = dielectric constant.
Field becomes 1/K times.
Energy stored becomes 1/K times as U= 1/2 C V^2 and V becomes 1/K times. The energy (k-1)/K * 1/2 Q^2/C is utilized in insertion of insulator between plates. in work done.
capacitance becomes K times. K = dielectric constant.
Field becomes 1/K times.
Energy stored becomes 1/K times as U= 1/2 C V^2 and V becomes 1/K times. The energy (k-1)/K * 1/2 Q^2/C is utilized in insertion of insulator between plates. in work done.
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