what is the ratio of the final energy to the initial energy, Uf/Ui, if the final state has been achieved by the following sequence of steps (starting from the initial situation)? Initial Area is 100 mm^2 and initial separation distance is 10 mm. the battery is disconnected the area is quadrupled the dielectric is taken out of the capacitor the separation is halved. Recreate the the table previously given below. Add a column to solve for the ratio.

what is the ratio of the final energy to the initial energy, Uf/Ui, if the final state has been achieved by the following sequence of steps (starting from the initial situation)? Initial Area is 100 mm^2 and initial separation distance is 10 mm. the battery is disconnected the area is quadrupled the dielectric is taken out of the capacitor the separation is halved. Recreate the the table previously given below. Add a column to solve for the ratio.

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Question

Predict: what is the ratio of the final energy to the initial energy, Uf/Ui, if the final state has been achieved by the following sequence of steps (starting from the initial situation)? Initial Area is 100 mm^2 and initial separation distance is 10 mm.

the battery is disconnected
the area is quadrupled
the dielectric is taken out of the capacitor
the separation is halved.

Recreate the the table previously given below. Add a column to solve for the ratio.

Ratio
Area
Distance
Dialectric Constant ..
Capacitance
...
mm^2
A
mm
d.
units
K
pF
1
100
10
0.443
..
2
400
1
0.708
...

Capacitance
Voltage
Charge
Electric Field
Potential Energy
..
...
...
...
...
ÞF
C
V
pC
V/d
pJ
U
1
p.443
1.49
0.66
149
0.492
...
2
p.708
0.932
0.66
186
0.308

Expert Solution

Step 1

Given,

Initial conditions,

Area, $A_{i} = 100 m m^{2}$

Separation, $d_{i} = 10 m m$

Dielectric constant, $k_{i} = 5$

Final conditions,

Area, $A_{f} = 400 m m^{2}$

Separation, $d_{f} = 5 m m$

Dielectric constant, $k_{f} = 1$

Capacitance,

$C = \frac{ε_{0} k A}{d}$

The energy is stored in the capacitor is given by,

$U = \frac{Q^{2}}{2 C} U = \frac{Q^{2}}{2 \frac{ε_{0} k A}{d}} U \propto \frac{d}{k A}$

Because the charge will not change on the plates.

So the ratio of final energy to initial energy,

$\frac{U_{f}}{U_{i}} = \frac{\frac{d_{f}}{k_{f} A_{f}}}{\frac{d_{i}}{k_{i} A_{i}}} \frac{U_{f}}{U_{i}} = \frac{\frac{5}{1 \times 400}}{\frac{10}{5 \times 100}} \frac{U_{f}}{U_{i}} = 0.625$