You have a parallel plate capacitor (with air between its plates) which is attac
ID: 1291629 • Letter: Y
Question
You have a parallel plate capacitor (with air between its plates) which is attached to a voltage source until the capacitor is fully charged.
(a)After detaching the capcitor from the voltage source, you move the capacitor's plates closer together; which of the following will occur (check all that apply)?
the capacitor's capacitance will increase
the capacitor's capacitance will decrease
the capacitor's capacitance will remain the same
the magnitude of the charge on each of the capacitor's plates will increase
the magnitude of the charge on each of the capacitor's plates will decrease
the magnitude of the charge on each of the capacitor's plates will remain constant
the potential difference across the capacitor's plates will increase
the potential difference across the capacitor's plates will decrease
the potential difference across the capacitor's plates will remain the same
the potential energy stored in the capacitor will increase
the potential energy stored in the capacitor will decrease
the potential energy stored in the capacitor will remain the same
more information is needed to be able to answer this
(b)While the capacitor is still attached to the voltage source, you move the capacitor's plates farther apart; which of the following will occur (check all that apply)?
the capacitor's capacitance will increase
the capacitor's capacitance will decrease
the capacitor's capacitance will remain the same
the magnitude of the charge on each of the capacitor's plates will increase
the magnitude of the charge on each of the capacitor's plates will decrease
the magnitude of the charge on each of the capacitor's plates will remain constant
the potential difference across the capacitor's plates will increase
the potential difference across the capacitor's plates will decrease
the potential difference across the capacitor's plates will remain the same
the potential energy stored in the capacitor will increase
the potential energy stored in the capacitor will decrease
the potential energy stored in the capacitor will remain
the same more information is needed to be able to answer this
Explanation / Answer
apply C = eoA/d and Q = CV and energy U = Q^2/2C
so here
when brouught closer:
the capacitor's capacitance will increase
the magnitude of the charge on each of the capacitor's plates will increase
the potential difference across the capacitor's plates will decrease
the potential energy stored in the capacitor will increase
b. when moved farther
the capacitor's capacitance will decrease
the magnitude of the charge on each of the capacitor's plates will decrease
the potential difference across the capacitor's plates will increase
the potential energy stored in the capacitor will decrease
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