Consider a parallel-plate capacitor having area A = 0.40 m 2 and separation d =

Question

Consider a parallel-plate capacitor having area A = 0.40 m2 and separation d = 0.010 m.

Consider a parallel-plate capacitor having area A = 0.40 m2 and separation d = 0.010 m. The capacitor is first attached to a battery having voltage 9 V. What are Q (in pC) and ?V? Now, the battery is removed, and a metallic slab of thickness a = 0.007 m is inserted. What are Q (in pC) and ?V? V After the battery is removed, what is the charge Q, and what is the capacitance C? Finally, the battery is reattached. What are Q (in pC) and ?V? pC Just as in part (A), what is the capacitance and the potential difference across the capacitor now?

Explanation / Answer

Q=CV

and C = eA/d

New C is 0.3 times of earlier so Q is also 0.3 times

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