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Chapter 24: Problem 33

A large parallel plate capacitor with plates that are square with side length \(1.00 \mathrm{~cm}\) and are separated by a distance of \(1.00 \mathrm{~mm}\) is dropped and damaged. Half of the areas of the two plates are pushed closer together to a distance of \(0.500 \mathrm{~mm}\). What is the capacitance of the damaged capacitor?

Short Answer

Expert verified
Answer: The capacitance of the damaged capacitor is \(13.28 \times 10^{-12} F\).

Step by step solution

01

Recall the capacitance formula for a parallel plate capacitor

The capacitance of a parallel plate capacitor is given by the formula: \(C = \epsilon_0 \frac{A}{d}\) where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (approximately \(8.85 \times 10^{-12} F/m\)), \(A\) is the area of one plate, and \(d\) is the distance between the plates.
02

Determine the area of the whole plate and half of the plate

We are given that the side length of the square plates is 1.00 cm. We'll first convert the side length to meters and then calculate the area of the whole plate and half of the plate: Side length = \(1.00 cm \times 10^{-2} m/cm = 0.0100 m\) Area of the whole plate = \((0.0100 m)^2 = 1.00 \times 10^{-4} \mathrm{~m^2}\) Area of half of the plate = \((1.00 \times 10^{-4} \mathrm{~m^2})/ 2 = 0.50 \times 10^{-4} \mathrm{~m^2}\)
03

Account for the different distances between the plates

The intact part of the capacitor still has a distance of 1.00 mm between the plates, while the damaged part has a distance of 0.500 mm. We will treat these as two separate capacitors connected in parallel and add their capacitances together. We need to convert the distances to meters: \(d_1\) = intact part distance = \(1.00 mm \times 10^{-3} m/mm = 0.00100 m\) \(d_2\) = damaged part distance = \(0.500 mm \times 10^{-3} m/mm = 0.000500 m\)
04

Calculate the capacitances of the intact and damaged parts

Using the capacitance formula, we will calculate the capacitances of the intact and damaged parts: \(C_1\) = \(\epsilon_0 \frac{A_1}{d_1} = (8.85 \times 10^{-12} F/m)(\frac{0.50 \times 10^{-4} \mathrm{~m^2}}{0.00100 m}) = 4.43 \times 10^{-12} F\) \(C_2\) = \(\epsilon_0 \frac{A_2}{d_2} = (8.85 \times 10^{-12} F/m)(\frac{0.50 \times 10^{-4} \mathrm{~m^2}}{0.000500 m}) = 8.85 \times 10^{-12} F\)
05

Add the capacitances together to find the total capacitance

Now we will add the capacitances of the intact and damaged parts to find the total capacitance: \(C_\text{total} = C_1 + C_2 = 4.43 \times 10^{-12} F + 8.85 \times 10^{-12} F = 13.28 \times 10^{-12} F\) Therefore, the capacitance of the damaged capacitor is \(13.28 \times 10^{-12} F\).

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Most popular questions from this chapter

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

A parallel plate capacitor consisting of a pair of rectangular plates, each measuring \(1.00 \mathrm{~cm}\) by \(10.0 \mathrm{~cm},\) with a separation between the plates of \(0.100 \mathrm{~mm},\) is charged by a power supply at a potential difference of \(1.00 \cdot 10^{3} \mathrm{~V}\). The power supply is then removed, and without being discharged, the capacitor is placed in a vertical position over a container holding de-ionized water, with the short sides of the plates in contact with the water, as shown in the figure. Using energy considerations, show that the water will rise between the plates. Neglecting other effects, determine the system of equations that can be used to calculate the height to which the water rises between the plates. You do not have to solve the system.

A proton traveling along the \(x\) -axis at a speed of \(1.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.0-\mathrm{cm}-\) wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.0 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

When working on a piece of equipment, electricians and electronics technicians sometimes attach a grounding wire to the equipment even after turning the device off and unplugging it. Why would they do this?
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