Physics: Properties of a Capacitor

[Phlame217]

A capacitor is charged and then removed from the circuit so that the charge remains constant. The distance between the plates is doubled. What happens to the following:
(Options are: Remain the same, Halved, Doubled, Quadrupled.)

Capacitance
Stored Energy
Electric Field Between Plates
Potential Difference between plates.

I believe that Capacitance is doubled, stored energy is is quadrupled, electric field is unchanged, and potential difference doubles. Is this correct?

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[BerkReviewTeach]

A capacitor is charged and then removed from the circuit so that the charge remains constant. The distance between the plates is doubled. What happens to the following:
(Options are: Remain the same, Halved, Doubled, Quadrupled.)

Capacitance
Stored Energy
Electric Field Between Plates
Potential Difference between plates.

I believe that Capacitance is doubled, stored energy is is quadrupled, electric field is unchanged, and potential difference doubles. Is this correct?

Capacitance:
If the plates are farther away from one another, then intuitively we can conclude they stabilize each other less, so it is less favorable for charge to be stored. Hence C decreases. If you think about taking the plates to infinite distance, then you'll note that eventaully C = 0.
Math tells us: C = Eo x A/d
So increasing d equates to decreasing C

Stored energy:
It takes work to pull the plates away from one another so we can assume the potential energy increases. How much depends on the equations.
Math tells us: Uelectrical = Qexp2/2C
Q doesn't change, so decreasing C equates to greater U. U should double according to this equation.

Electric Field:
Fields dissipate with distance