An empty parallel plate capacitor is connected between theterminals of a 11.8-V

Question

An empty parallel plate capacitor is connected between theterminals of a 11.8-V battery and charges up. The capacitor is thendisconnected from the battery, and the spacing between thecapacitor plates is doubled. As a result of this change, what isthe new voltage between the plates of the capacitor?
I do not know how to do this,could you please help me?
I do not know how to do this,could you please help me?

Explanation / Answer

the charge in capacitor is unchanged => Q = Vo*Co =Vo*o*A/d when double the space between 2 plates ,the new capacitance : C =o*A/(2d) = Co/2 => the neew voltage : V = Q/C = Vo*Co/(Co/2) = 2Vo = 2*15.6= 31.2 V

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