A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:
Correct answer: C. Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
- A. Reduction of charge on the plates and increase of potential difference across the plates
- B. Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
- C. Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
- D. None of the above
Explanation
Since the battery is disconnected, the charge (Q) on the plates remains constant. Inserting a dielectric increases the capacitance (C), and since V=Q/C, the potential difference (V) decreases. The stored energy U = Q²/(2C) also decreases.
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About Electrostatics
Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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