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A parallel plate capacitor is charged by...

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab having dielectric constant (relative permittivity) K, is now introduced between its two plates in order to occupy the space completely. State, in terms of K, its effect on the following:
The capacitance of the capacitor

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To solve the problem, we need to analyze the effect of introducing a dielectric slab with dielectric constant \( K \) into a parallel plate capacitor that has already been charged and disconnected from the battery. ### Step-by-Step Solution: 1. **Understand the Initial Capacitance**: The capacitance \( C \) of a parallel plate capacitor without a dielectric is given by the formula: \[ C = \frac{\varepsilon_0 A}{d} ...
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