Question

Suppose two capacitor plates have an area of 0.0300 m2 and are initially separated by 1.00 mm. One plate holds 2.30 μC of charge, while the other plate holds -2.30 μC of charge. Part A How much energy is required to increase the plate separation to 2.00 mm?

Answer 1

Concept - find the initial capacitance and use the charge to find the initial energy. Find the final capacitance after the separation is doubled. Find the final energy using the charge which remains the same. Then change in energy is the energy required.

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Answer 2

To calculate the energy required to increase the plate separation, we can use the formula for the energy stored in a capacitor:

E = (1/2) * C * V^2

Where: E is the energy stored in the capacitor C is the capacitance of the capacitor V is the potential difference (voltage) across the capacitor

First, let's calculate the initial capacitance of the capacitor:

C = ε₀ * (A / d)

Where: ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m) A is the area of the capacitor plates (0.0300 m^2) d is the initial separation between the plates (1.00 mm or 0.001 m)

C = (8.85 x 10^-12 F/m) * (0.0300 m^2 / 0.001 m) = 2.655 x 10^-10 F

Next, we need to calculate the initial potential difference (V) across the capacitor. The potential difference can be obtained by dividing the charge (Q) on one of the plates by the capacitance (C):

V = Q / C

Since one plate holds 2.30 μC of charge, the potential difference is:

V = (2.30 x 10^-6 C) / (2.655 x 10^-10 F) = 8662.36 V

Now, let's calculate the final capacitance of the capacitor when the plate separation increases to 2.00 mm or 0.002 m:

C' = ε₀ * (A / d')

Where: d' is the final separation between the plates (0.002 m)

C' = (8.85 x 10^-12 F/m) * (0.0300 m^2 / 0.002 m) = 1.3275 x 10^-10 F

To calculate the final potential difference (V') across the capacitor, we can use the equation:

V' = Q / C'

Since the charge (Q) on the plates remains the same, V' will be the same as V:

V' = V = 8662.36 V

Finally, let's calculate the energy (E') required to increase the plate separation:

E' = (1/2) * C' * V'^2

E' = (1/2) * (1.3275 x 10^-10 F) * (8662.36 V)^2

E' = 5.585 x 10^-3 J or 5.585 mJ

Therefore, the energy required to increase the plate separation from 1.00 mm to 2.00 mm is approximately 5.585 millijoules