Physics · Capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor
A parallel plate condenser with plate separation d is charged with the help of b
A parallel plate condenser with plate separation d is charged with the help of battery so that \( V_{0} \) energy is stored in the system. The battery is now removed. a plate of dielectric constant k and thickness d is placed between the plates of condenser. The new energy of the system will be :
- A. \( V_{0} k^{-2} \)
- B. \( k^{2} V_{0} \)
- C. \( V_{0} k^{-1} \)
- D. \( k V_{0} \)
Step-by-step solution
Initially, energy stored V0 = Q²/(2C). After battery removal, charge Q remains constant. Inserting a dielectric of constant k and thickness d fills the entire gap, increasing capacitance to C' = kC. New energy U' = Q²/(2C') = Q²/(2kC) = (1/k) * (Q²/(2C)) = V0/k = V0 k^{-1}.
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