Physics · Capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor
A capacitor of capacitance is charged by a battery. Now the space between the pl
A capacitor of \( 10 \mu F \) capacitance is charged by a \( 12 V \) battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant \( K=3 \) and again it is charged. The magnitude of the charge is :
- A. \( 120 \mu C \)
- B. \( 240 \mu C \)
- C. \( 360 \mu C \)
- D. \( 480 \mu C \)
Step-by-step solution
Initial charge: Q0 = C0V = 10 μF × 12 V = 120 μC. After inserting dielectric with K=3, capacitance becomes C = K C0 = 30 μF. Recharging with same 12 V battery gives charge Q = CV = 30 μF × 12 V = 360 μC.
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