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A 10.0 -\muF parallel-plate capacitor is connected to a 12.0 -V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. (a) A voltmeter is connected across the two plates without discharging them. What does it read? (b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

Short Answer

Expert verified
(a) 12.0 V; (b-i) 24.0 V; (b-ii) 3.0 V.

Step by step solution

01

Determine the Initial Charge on the Capacitor

The initial charge \( Q \) on a capacitor is given by the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the voltage. For our capacitor, \( C = 10.0 \times 10^{-6} \, F \) and \( V = 12.0 \, V \). Therefore, the initial charge \( Q \) is: \[ Q = 10.0 \times 10^{-6} \times 12.0 = 1.2 \times 10^{-4} \, C \].
02

Voltage Reading Across the Capacitor

Once the battery is disconnected, the charge \( Q \) on the capacitor remains constant. The voltage \( V \) across a capacitor is given by \( V = \frac{Q}{C} \). Therefore, the voltmeter will read the original voltage: \[ V = \frac{1.2 \times 10^{-4}}{10.0 \times 10^{-6}} = 12.0 \, V \]. So the voltmeter initially reads 12.0 V.
03

Effect of Doubling Plate Separation

When the plate separation \( d \) is doubled, the capacitance \( C \) of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_0 A}{d} \). Thus, doubling \( d \) halves \( C \). The new capacitance is \( C' = \frac{C}{2} \). Using \( V' = \frac{Q}{C'} \), the new voltage is: \[ V' = \frac{1.2 \times 10^{-4}}{\frac{10.0 \times 10^{-6}}{2}} = 24.0 \, V \]. Therefore, the voltmeter would read 24.0 V if the plate separation is doubled.
04

Effect of Doubling the Radius of Each Plate

Doubling the radius of the plates increases the area \( A \) by a factor of four (since area is proportional to the square of the radius), yielding \( C'' = 4C \). The voltage \( V'' = \frac{Q}{C''} \) is then: \[ V'' = \frac{1.2 \times 10^{-4}}{4 \times 10.0 \times 10^{-6}} = 3.0 \, V \]. Hence, the voltmeter would read 3.0 V if the radius of each plate is doubled.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel-Plate Capacitor
Parallel-plate capacitors are fundamental devices in electronics, consisting of two conductive plates separated by an insulating material known as a dielectric. Both plates are parallel to each other, which is why they are aptly named parallel-plate capacitors. Capacitors are widely used to store electrical energy in the form of an electric field set up between these plates.
The key components of a parallel-plate capacitor include:
  • **Two Conductive Plates**: Typically made of metal, these plates hold the electrical charge.
  • **Dielectric Material**: This insulating layer separates the plates and improves the capacitor's ability to store charge, known as its capacitance.
  • **Electric Field**: Once charged, an electric field is established between the plates, allowing for energy storage.
The distance between the plates and the surface area of the plates itself significantly affect the overall capacitance of the capacitor. These factors play a crucial role in how much charge the capacitor can store.
Capacitance Formula
Capacitance is a fundamental property of capacitors, defining their ability to store charge per unit voltage. The standard capacitance formula for a parallel-plate capacitor is given by:\[ C = \frac{\varepsilon_0 A}{d} \]where:
  • **\(C\)** is the capacitance,
  • **\(\varepsilon_0\)** (epsilon naught) is the permittivity of free space, a constant representing the ability of the vacuum to allow electric field lines to pass through it.
  • **\(A\)** is the area of one of the plates.
  • **\(d\)** is the separation between the plates.
The formula illustrates that as the plate area \(A\) increases, the capacitance also increases. Conversely, as the distance \(d\) between the plates increases, the capacitance diminishes. This relationship highlights why capacitors with large plate areas and small gaps hold more charge, making them suitable for various electronic applications. It becomes clear that alterations to the structure, such as changing the plate separation or plate size, directly affect capacitance.
Voltage and Charge
The concepts of voltage and charge are closely intertwined when discussing capacitors. Once a capacitor is connected to a battery, it begins accumulating charge until it reaches its full capacity. The amount of charge \(Q\) stored in a capacitor can be calculated using the formula:\[ Q = C \times V \]where:
  • **\(Q\)** is the charge stored,
  • **\(C\)** is the capacitance of the capacitor,
  • **\(V\)** is the voltage applied across the capacitor.
In practical terms, if a capacitor is fully charged and then disconnected from the voltage source, it will retain its charge. This means that the voltage across its plates remains consistent until discharge starts through another path. For instance, if you alter the physical characteristics of the capacitor, such as doubling the separation between the plates, it affects the voltage necessary to maintain the charge. This is why, without losing charge, disconnecting the battery but doubling the plate separation might lead to a higher voltage reading as indicated by the voltaic measurements, since the capacitance is halved and thereby, the voltage doubles. This demonstrates the importance of understanding how modifications to a capacitor's structure affect its performance.

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Most popular questions from this chapter

A capacitor is made from two hollow, coaxial copper cylinders, one inside the other. There is air in the space between the cylinders. The inner cylinder has net positive charge and the outer cylinder has net negative charge. The inner cylinder has radius \(2.50 \mathrm{mm},\) the outer cylinder has radius \(3.10 \mathrm{mm},\) and the length of each cylinder is 36.0 \(\mathrm{cm} .\) If the potential difference between the surfaces of the two cylinders is \(80.0 \mathrm{V},\) what is the magnitude of the electric field at a point between the two cylinders that is a distance of 2.80 \(\mathrm{mm}\) from their common axis and midway between the ends of the cylinders?

B10 Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{\boldsymbol{E}}\) in the wall-between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) Atypical cell in the human body has a volume of\(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

A parallel-plate air capacitor is made by using two plates 16 \(\mathrm{cm}\) square, spaced 3.7 \(\mathrm{mm}\) apart. It is connected to a \(12-\mathrm{V}\) battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(7.4 \mathrm{mm},\) what are the answers to parts (a)-(d)?

A parallel-plate capacitor has capacitance \(C=12.5 \mathrm{pF}\) when the volume between the plates is filled with air. The plates are circular, with radius 3.00 \(\mathrm{cm}\) . The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 \(\mathrm{pC}\) (a) What is the dielectric constant \(K\) of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

A cylindrical capacitor has an inner conductor of radius 1.5 \(\mathrm{mm}\) and an outer conductor of radius 3.5 \(\mathrm{mm}\) . The two conductors are separated by vacuum, and the entire capacitor is 2.8 \(\mathrm{m}\) long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 \(\mathrm{mV}\) higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

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