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Chapter 17: Problem 76

A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$

Short Answer

Expert verified
Question: Calculate the work done on a parallel plate capacitor when its plate separation is increased from 1.00 cm to 2.00 cm while keeping the voltage constant at 20 V. The area of the plates is 314 cm虏 and the vacuum permittivity is 8.85 脳 10鈦宦孤� F/m. Answer: The work done on the capacitor is -2.774 脳 10鈦烩伖 J.

Step by step solution

01

Find the initial capacitance

The formula for capacitance of a parallel plate capacitor is: \(C = \frac{\epsilon_0 A}{d}\) Where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity, \(A\) is the area of the plates, and \(d\) is the distance between the plates. Using the given values, we can find the initial capacitance. \(\epsilon_0 = 8.85 脳 10^{-12} \mathrm{F/m}\) \(A = 314 \mathrm{cm}^2 \times 10^{-4} \mathrm{m}^2/\mathrm{cm}^2 = 0.0314\,\mathrm{m}^2\) \(d_1 = 1.00\, \mathrm{cm} \times 10^{-2} \mathrm{m/cm} = 0.010\,\mathrm{m}\) \(C_1 = \frac{\epsilon_0 A}{d_1} = \frac{8.85 脳 10^{-12} \mathrm{F/m} \times 0.0314\,\mathrm{m}^2}{0.010\,\mathrm{m}} = 2.774 \times 10^{-11}\,\mathrm{F}\)
02

Find the final capacitance

Now, we can find the final capacitance when the plate separation is increased to \(2.00\,\mathrm{cm}\). \(d_2 = 2.00\, \mathrm{cm} \times 10^{-2} \mathrm{m/cm} = 0.020\,\mathrm{m}\) \(C_2 = \frac{\epsilon_0 A}{d_2} = \frac{8.85 脳 10^{-12} \mathrm{F/m} \times 0.0314\,\mathrm{m}^2}{0.020\,\mathrm{m}} = 1.387 \times 10^{-11}\,\mathrm{F}\)
03

Calculate the energy stored in the capacitor before and after

The energy stored in a capacitor can be calculated using the formula: \(U = \frac{1}{2}CV^2\) We can calculate the energy stored in the capacitor initially and after the plate separation is increased: \(U_1 = \frac{1}{2}C_1V^2 = \frac{1}{2}(2.774 \times 10^{-11}\,\mathrm{F})(20.0\,\mathrm{V})^2= 5.548 \times 10^{-9}\,\mathrm{J}\) \(U_2 = \frac{1}{2}C_2V^2 = \frac{1}{2}(1.387 \times 10^{-11}\,\mathrm{F})(20.0\,\mathrm{V})^2= 2.774 \times 10^{-9}\,\mathrm{J}\)
04

Calculate the work done on the capacitor

The work done on the capacitor can be found by taking the difference between the final and initial energies stored in the capacitor: \(W = U_2 - U_1 = 2.774 \times 10^{-9}\,\mathrm{J} - 5.548 \times 10^{-9}\,\mathrm{J} = -2.774 \times 10^{-9}\,\mathrm{J}\) The negative sign indicates that work is done by the capacitor, meaning that the energy is transferred from the capacitor to an external system. This is the work done on the capacitor as the plate separation is increased to \(2.00\,\mathrm{cm}\).

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Most popular questions from this chapter

To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?
A hydrogen atom has a single proton at its center and a single electron at a distance of approximately \(0.0529 \mathrm{nm}\) from the proton. (a) What is the electric potential energy in joules? (b) What is the significance of the sign of the answer?
Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?
A 200.0 - \(\mu\) F capacitor is placed across a \(12.0-\mathrm{V}\) battery. When a switch is thrown, the battery is removed from the capacitor and the capacitor is connected across a heater that is immersed in $1.00 \mathrm{cm}^{3}$ of water. Assuming that all the energy from the capacitor is delivered to the water, what is the temperature change of the water?
Figure 17.31 b shows a thundercloud before a lightning strike has occurred. The bottom of the thundercloud and the Earth's surface might be modeled as a charged parallel plate capacitor. The base of the cloud, which is roughly parallel to the Earth's surface, serves as the negative plate and the region of Earth's surface under the cloud serves as the positive plate. The separation between the cloud base and the Earth's surface is small compared to the length of the cloud. (a) Find the capacitance for a thundercloud of base dimensions \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\) located \(550 \mathrm{m}\) above the Earth's surface. (b) Find the energy stored in this capacitor if the charge magnitude is \(18 \mathrm{C}\).
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