Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates, as shown. Potential difference between the plates after the dielectric slab has been introduced is:
Correct answer: C. 11.5 V
- A. 18.5 V
- B. 13.5 V
- C. 11.5 V
- D. 6.5 V
Explanation
When a dielectric slab of thickness 3 mm and dielectric constant K = 10 is inserted between the plates of a capacitor, the effective separation between the plates changes. The potential difference across the capacitor is given by the relation V = Q/C, where C is the capacitance. The introduction of the dielectric increases the capacitance, which is inversely proportional to the potential difference, thus reducing it. The correct answer is 11.5 V, as calculated by considering the series combination of the air gap and the dielectric slab.
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