Physics, 03.07.2019 20:00 tmanpierce

A1.97-pf capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-v battery and fully charged. then the separation between the plates is adjusted so that the energy stored in the capacitor is increased by a factor of 3.5. what is the new plate separation distance?

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Answer from: Ajaxr2432

If the plate separation is adjusted after disconnecting the battery, the new plate separation is 9.21 mm

If the plate separation is adjusted with the battery connected, the new plate separation is 0.11 mm

The capacitance of an air filed parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, $\epsilon_0$ is the permittivity of free space , A is the area of the plates and D is the plate separation.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Thus if the pate separation is changed from d₁ to d₂ , then the ratio of capacitances of the capacitor in the two cases is given by,

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Case (i)

The capacitor is disconnected from the battery after being charged fully and then the plate separation is changed. In such a case, the charge on the plates of the capacitor remains constant, while the capacitance changes.

The initial energy E₁ stored in the capacitor is given by,

$E_1=\frac{Q^2}{2C_1}$ ......(3)

After the plate separation changes to d₂ , the capacitance changes to C₂, but the charge Q remains the same.

Therefore,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

Divide equation (4) by (3)

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

The energy is increased by a factor of 3.5.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Case (2)

If the capacitor remains connected to the battery, the potential difference V between the plates remains the same.

The initial energy is given by,

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy after the plate separation changes to d₂ is given by,

$E_2=\frac{1}{2} C_2V^2$ .....(6)

From equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this case,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

If the plate separation is adjusted after disconnecting the battery, the new plate separation is 9.21 mm

If the plate separation is adjusted with the battery connected, the new plate separation is 0.11 mm

Answer from: Quest

grav potential energy

Answer from: Quest

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