A parallel plate capacitor is charged to voltage V and then disconnected from th

Question

A parallel plate capacitor is charged to voltage V and then disconnected from the battery. Person A says that the voltage will decrease if the plates are pulled apart. Person B says that the voltage will remain the same. Which one, if either, is correct, and why? Person B because the maximum voltage is determined by the battery Person B, because the charge per unit area on the plates does not change. Person A, because charge is transferred from one plate to the other when the plates are separated. Person A, because the force each plate exerts on the other decreases when the plates are pulled apart. Neither, because the voltage increases when the plates are pulled apart.

Explanation / Answer

Q3. when fully charged, charge on capacitor plates=Q=C*V

when battery is disconnected, charge remains constant .

when plates are pulled apart, as capacitance=epsilon*A/d

where A=area and d=distance between plates

d increases and capacitance decreases

as charge remains constant, voltage difference=charge/capacitance increases

so none of them are correct.

option e is the correct answer.

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