If a dielectric is inserted between the plates of an isolated charged air capacitor, the energy of the capacitor willIncreaseDecreaseRemains unchangedFirst increases and then decreases
Question
If a dielectric is inserted between the plates of an isolated charged air capacitor, the energy of the capacitor willIncreaseDecreaseRemains unchangedFirst increases and then decreases
Solution
The energy of the capacitor decreases.
Here's why:
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An isolated charged air capacitor means that the capacitor is not connected to any battery or power source, and it has some charge stored in it.
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When a dielectric is inserted between the plates of the capacitor, the capacitance of the capacitor increases. This is because the dielectric increases the ability of the capacitor to store charge.
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The energy stored in a capacitor is given by the formula U = 1/2 * C * V^2, where U is the energy, C is the capacitance, and V is the voltage.
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Since the capacitor is isolated, the charge Q on the capacitor remains constant. The relationship between charge, capacitance, and voltage is Q = C * V. Therefore, if the capacitance C increases, the voltage V must decrease to keep the charge Q constant.
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Substituting the decreased voltage back into the energy formula, we see that the energy U of the capacitor decreases. This is because the energy is proportional to the square of the voltage, so a decrease in voltage results in a decrease in energy.
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