A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correctincrease in the stored energydecrease in the potential differencedecrease in the electric fieldincrease in the capacitance
Question
A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correctincrease in the stored energydecrease in the potential differencedecrease in the electric fieldincrease in the capacitance
Solution
The statement that is not correct is "increase in the stored energy".
Here's why:
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Increase in the capacitance: When a dielectric is inserted between the plates of a capacitor, it increases the capacitance. The dielectric reduces the electric field between the plates, allowing more charge to be stored for the same voltage. Hence, the capacitance (C = Q/V) increases.
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Decrease in the potential difference: Since the capacitance increases and the charge remains constant (because the battery is disconnected), the potential difference (V = Q/C) decreases.
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Decrease in the electric field: The electric field in a capacitor is given by E = V/d, where d is the distance between the plates. When a dielectric is inserted, the potential difference V decreases, hence the electric field also decreases.
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Increase in the stored energy: This is the incorrect statement. The energy stored in a capacitor is given by U = 1/2 * C * V^2. When a dielectric is inserted, the capacitance C increases and the potential difference V decreases. However, since the potential difference is squared in this formula, the overall effect is a decrease in the stored energy, not an increase.
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