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A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased. What happens to the potential difference between the plates

Sagot :

Answer:

V = E * d      voltage equals charge on plates * separation

F = E q     force on charge q equals electric field * charge

W = F * d     work done equals force times the distance moved

From the very first equation, the voltage depends on the distance between the plates - while the electric field remains constant

The potential difference will decrease as the separation of the plates increases because W remains constant - work done in moving test charge from one plate to another is constant - voltage is constant

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