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# Answer to Question #167407 in Electrical Engineering for jeevan

Question #167407

A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed.

a) If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement?

b) If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?

1
2021-03-01T02:15:48-0500

Q167407

A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed.

a) If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement?

Solution :

Let us assume that the mica sheet is of the same thickness as the distance between the two plates.

The area of the mica sheet is 3.00 cm2.

Area of parallel plate capacitor = 5.00 cm2.

After placing this mica sheet there would be some are uncovered by mica sheet.

This uncovered space between the plates will have the dielectric constant of the air.

Uncovered area = 5.00 cm2 - 3.00 cm2 = 2.00 cm2 .

Step 1: Find the capacitance between the uncovered plates and the plates covered with a mica sheet.

the initial area of capacitance with air between the plates was 5.00 cm2.

Now the area of capacitance with air between the plates = 2.00 cm2.

Consider the formula, "C = \\frac{ \u03b5_0*A}{d }"

ε0 is the permittivity of vacuum, A is the area of the plate, d is the distance between the plate.

We can see that capacitance is directly related to the Area of the plate.

So the capacitance of the uncovered plate will be

"C_{ uncovered} = \\frac{3.50 *10^{-12 }F * 2.00cm2}{5.00cm^2} = 1.4*10^{-12} F"

the relative permittivity of the mica sheet is not given in the question.

The sheet of Mica with εr=3.5 is placed between the plates,

So the capacitance between the mica sheet will be

"C_{mica} = \u03b5r *C_{air} = 3.5 * 3.50 * 10^{-12} F = 1.225 * 10^{-11} F"

This will be the capacitance when the whole space between the plates is replaced by the mica sheet. But since only 3.00 cm2 space out of 5.00 cm2 is replaced by a mica sheet, so the

capacitance due to mica sheet will be

"C_{mica} = 1.225 * 10^{-11} F * \\frac{3cm^{2}}{5.00cm^{2} } = 7.35 * 10^{-12} F"

Step 2: To find the total capacitance

the 2.00cm2 uncovered space and 3.00 cm2 plates covered with mica will act like there are two capacitances in parallel. So the resultant capacitance will be sum of their capacitance

"C_{total } = C _{covered \\space with \\space mica\\space sheet} + C _ { uncovered (air )}"

"= 1.4*10^{-12} F + 7.35 * 10^{-12} F"

or 8.75 nF

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b) If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?

Solution :

Use the formula , C = ε0 * A / d .

where, ε0 = 8.85 * 10-12 F/m . A is the area of the plate in m2 and d is the distance between the plate in 'm' .

plug A = 5.00 * 10-4 m2 , and C = 3.50 * 10-12 F , we will get

d = 0.001264 m.

After moving the plate 1.00 mm = 0.001 meter farther the new distance between the plates is

d new = 0.001264 + 0.001 = 0.002264 m.

Again use the same formula, C new = ε0 * A / dnew

and find the new capacitance

Cnew = 8.85 * 10^-12 * 5.00 * 10^-4 / 0.002264 m

= 1.95 * 10--12 F .

The charge, Q = C * V = 3.50 * 10^-12 F * 12.0 = 4.2 *10^-11 coulomb .

Work done = "\\frac{Q^2}{2C_{final} } - \\frac{Q^2}{2C_{initial}} = \\frac{(4.2*10^{-11})^2}{2* 1.95*10^{-12} } - \\frac{(4.2*10^{-11})^2}{2*3.50*10^{-12}}"

= 4.77 Joules

Hence the work done in moving the plates will be 4.77 Joules.

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