A parallel plate capacitor is completely charged and battery is detached from it. Then adielectric slab is inserted between its parallel plates.Which of the following options is CORRECT with respect to potential difference and energy?
Correct answer: D. The potential difference across the capacitor decreases and the stored energy increases.
- A. The potential difference across the capacitor increases and the stored energy remains the same.
- B. The potential difference across the capacitor decreases and the stored energy decreases.
- C. The potential difference across the capacitor remains the same and the stored energy remains the same.
- D. The potential difference across the capacitor decreases and the stored energy increases.
Explanation
When a dielectric slab is inserted into a charged capacitor that is isolated (battery detached), the potential difference across the capacitor decreases because the dielectric reduces the electric field within it. The charge on the capacitor remains constant because there is no source to provide additional charge or remove any charge.The energy stored in the capacitor can be expressed as U = (1/2)CV2. The capacitance C increases due to the dielectric, and since Q = CV (where Q is constant), the potential difference V decreases. However, the increase in capacitance more than compensates for the decrease in V squared, leading to an overall increase in the stored energy.Options A and B are incorrect because they misunderstand the effect of the dielectric on potential difference and stored energy. Option C is incorrect because it incorrectly assumes that neither potential difference nor stored energy changes.
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About Capacitors
Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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