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dartz3429 dartz3429
wrote...
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12 years ago
Suppose a parallel-plate capacitor is connected to a battery and charged so the potential difference between the plates of the capacitor is equal to the potential difference between the terminals of the battery. If the plates are now moved farther apart (while the capacitor is still connected to the battery):
The charge will decrease.
WHY?! (Please explain in terms of equations/formulas)...
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wrote...
12 years ago
There are two formulas to explain this.

Q = C * V or Charge = Capacitance times the Potential Difference (given in volts by the battery)

The V will not change due to the distance but the C will. C = k * E * A all divided by D.

k is a constant known as a dielectric to increase the ability to hold charge
E is a constant (permittivity of free space = 8.85 * 10^-12)
A is area of the capacitor
D is the distance between the plates, as D increases the C value decreases and since V is constant, your charge Q is decreased.

hope that helped
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