A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0U0U_0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is UUU. Find the ratio U/U0U/U0.

Answer :

elcharly64

Answer:

[tex]\displaystyle \frac{U}{U_o}=\frac{1}{2}[/tex]

Explanation:

Energy Stored in Capacitors

The capacitance of a parallel-plate capacitor is given by

[tex]\displaystyle C=\epsilon_0\frac{A}{d}[/tex]

Where [tex]\epsilon_o[/tex] is the permitivity of the dielectric, A is the area of the plates and d is their separation. If the separation was doubled, the new capacitance would be

[tex]\displaystyle C'=\epsilon_0\frac{A}{2d}=\frac{C}{2}[/tex]

The energy stored in the initial condition is

[tex]\displaystyle U_o=\frac{CV^2}{2}[/tex]

And the energy stored when the separation doubles is

[tex]\displaystyle U=\frac{CV^2}{4}[/tex]

Thus the ratio

[tex]\displaystyle \frac{U}{U_o}=\frac{\frac{CV^2}{4}}{\frac{CV^2}{2}}=\frac{1}{2}[/tex]

[tex]\boxed{\displaystyle \frac{U}{U_o}=\frac{1}{2}}[/tex]

The energy is half the initial energy

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