Question 2 A capacitor with plates separated by 0.0260 m is charged to a potential difference of 6.50 V. All wires and batteries are then disconnected, and the two plates are pulled apart to a new separation distance of 0.0520 m. What is the potential difference after the increase in separation? O 3.25 V O 13.0 V O 26.0 V O 6.50 V
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The capacitance of a parallel plate capacitor is given by: $$C = \frac{\epsilon_0 A}{d}$$ where $\epsilon_0$ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. Show more…
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Madhur L.
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A capacitor with plates separated by 0.0180 m is charged to a potential difference of 7.50 V. All wires and batteries are then disconnected, and the two plates are pulled apart to a new separation distance of 0.0360 m. If the initial charge on the capacitor is 8.00 nC, what is the charge after the increase in separation? 16.0 nC 32.0 nC 8.00 nC 4.00 nC
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