A parallel plate air capacitor is charged to apotential difference of ...
If we increase the distance between the plates its capacity decreases resulting in higher potential as we know Q = CV. Since Q is constant (battery has been disconnected), on decreasing C, V will increase
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A parallel plate air capacitor is charged to apotential difference of ...
Understanding Capacitors
A parallel plate capacitor consists of two conductive plates separated by a certain distance, with an insulating material (dielectric) in between. When connected to a battery, it stores charge (Q) and develops a potential difference (V) across its plates.
Charging and Disconnecting
- When a capacitor is connected to a battery, it charges up to a voltage V.
- Upon disconnecting the battery, the charge (Q) on the plates remains constant because there is no external circuit to allow charge to flow away.
Effect of Increasing Plate Distance
- When the distance (d) between the plates is increased using an insulating handle, the capacitance (C) of the capacitor changes.
- The capacitance of a parallel plate capacitor is given by the formula C = ε0 * A / d, where ε0 is the permittivity of free space and A is the area of the plates.
Capacitance and Voltage Relationship
- Since the capacitor is isolated (disconnected from the battery), the charge (Q) remains constant.
- The relationship between charge (Q), capacitance (C), and voltage (V) is given by Q = C * V.
- When distance (d) increases, capacitance (C) decreases, which leads to an increase in voltage (V) since Q is constant.
Conclusion
- Therefore, as the distance between the plates increases, the potential difference (V) across the plates increases, confirming that the correct answer is option 'C': the potential difference increases.
This behavior is a fundamental characteristic of capacitors and highlights the interplay between charge, capacitance, and voltage.