What happens to charge when the plate separation is increased, and why?
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The question is ambiguous. Is the capacitor still connected to a voltage source or is it charged and isolated?
If isolated, then there is nowhere for the charge to go and so it remains constant as the plates are separated. Since the charge is constant and the capacitance is decreasing with increased separation the voltage must increase . This makes sense as separating the plates involves pulling them apart against the attraction of the charged plates for each other.
If the battery remains connected as the plates pull apart, then charge is driven ( by mutual repulsion ) back into the battery; the capacitance drops, the voltage stays the same and the charge decreases.
Field strength is volts per metre.... again, if isolated , the voltage increases, while the distance increases so the filed strength stays constant.
If the battery is connected, the voltage stays constant and the distance increases so the Field strength decreases.
If isolated, then there is nowhere for the charge to go and so it remains constant as the plates are separated. Since the charge is constant and the capacitance is decreasing with increased separation the voltage must increase . This makes sense as separating the plates involves pulling them apart against the attraction of the charged plates for each other.
If the battery remains connected as the plates pull apart, then charge is driven ( by mutual repulsion ) back into the battery; the capacitance drops, the voltage stays the same and the charge decreases.
Field strength is volts per metre.... again, if isolated , the voltage increases, while the distance increases so the filed strength stays constant.
If the battery is connected, the voltage stays constant and the distance increases so the Field strength decreases.
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