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A capacitor plates are charged by a battery with ‘V’ volts. After charging, battery is disconnected and a dielectric slab with dielectric constant ‘K’ is inserted between its plates, the potential across the plates of a capacitor will become:
- a)Zero
- b)V/2
- c)V/K
- d)KV
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A capacitor plates are charged by a battery with ‘V’ volts...
Battery is disconnected.
Q = Charge remains context C′ = KC
Q′ = C′V′
Q = C′V′
Q = KCV′
V′ = 
Most Upvoted Answer
A capacitor plates are charged by a battery with ‘V’ volts...
Effect of Dielectric Insertion on Capacitor Potential
The potential across the plates of a capacitor changes when a dielectric slab is inserted between the plates. Here's how the potential changes:
Dielectric Constant
- The dielectric constant, denoted by 'K', is a measure of how much the capacitance of a capacitor increases when a dielectric material is inserted between its plates.
- The dielectric constant 'K' is greater than 1 for all dielectric materials.
Effect on Potential
- When the dielectric slab with dielectric constant 'K' is inserted between the plates of the capacitor, the potential across the plates decreases.
- The potential across the plates of the capacitor becomes V/K, where 'V' is the initial potential across the plates before the dielectric insertion.
Explanation
- Initially, the potential across the plates is 'V' volts when the plates are charged by a battery.
- When the dielectric slab is inserted, the electric field between the plates decreases due to the polarization of the dielectric material.
- The potential difference across the plates decreases by a factor of 'K' due to the presence of the dielectric material with dielectric constant 'K'.
- Therefore, the new potential across the plates of the capacitor becomes V/K after the dielectric insertion.
In conclusion, the potential across the plates of a capacitor decreases to V/K when a dielectric slab with dielectric constant 'K' is inserted between the plates.
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