A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is
A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is
- A
Charge $Q$
- B
Potential $V$
- C
Capacity $C$
- D
Energy $U$
Similar Questions
A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is
A parallel plate capacitor has plates with area $A$ and separation $d$ . A battery charges the plates to a potential difference $V_0$ . The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced, is
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?
Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. A dielectric slab of dielectric constant $4.0$ is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively
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Write the formula of capacitance of capacitor having dielectric constant $\mathrm{K} = 2$.
Write the formula of capacitance of capacitor having dielectric constant $\mathrm{K} = 2$.