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Chapter 18: Problem 41

A 10.0\(\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 \(\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

Short Answer

Expert verified
(a) 1.2 mC, (b) 0.6 mC, (c) 4.8 mC.

Step by step solution

01

Identify the Known Values

The problem gives us a parallel-plate capacitor with a capacitance of 10.0 \(\mu \mathrm{F}\) and a battery with a voltage of 12.0 \(\mathrm{V}\). We are asked to calculate charges on the plates under different conditions.
02

Use Capacitance Formula (a)

To find the charge on the capacitor plates, use the formula for charge \(Q\) in a capacitor: \(Q = C \times V\), where \(C\) is the capacitance and \(V\) is the voltage. Here, \(C = 10.0 \times 10^{-6} \mathrm{F}\) and \(V = 12.0 \mathrm{V}\). Substitute these values into the formula:\[Q = 10.0 \times 10^{-6} \times 12.0 = 1.2 \times 10^{-4} \mathrm{C}\]The charge on each plate is 1.2 \(\mathrm{mC}\).
03

Consider Charge with Separation Doubled (b)

If the plate separation is doubled, the capacitance \(C\) is halved because \(C\) is inversely proportional to the separation for parallel plates. But since the battery remains connected, the voltage \(V = 12.0 \mathrm{V}\) does not change. Charge \(Q\) is still \(C \times V\). As capacitance halves, the charge halves:\[Q = \frac{1}{2} \times 1.2 \times 10^{-4} = 0.6 \times 10^{-4} \mathrm{C}\]The charge becomes 0.6 \(\mathrm{mC}\).
04

Consider Charge with Radius Doubled (c)

When the radius of the plates is doubled, the area of the plates increases by a factor of four \((A \propto r^2)\). The capacitance \(C\) is directly proportional to the area, so capacitance increases by four times. The voltage \(V = 12.0 \mathrm{V}\) remains the same. Thus, the charge also increases by four times:\[Q = 4 \times 1.2 \times 10^{-4} = 4.8 \times 10^{-4} \mathrm{C}\]The charge becomes 4.8 \(\mathrm{mC}\).

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Key Concepts

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

Capacitance and Voltage Relationship
Capacitance is a property of a capacitor that indicates how much charge it can hold per unit of voltage applied across its plates. Specifically, the charge \( Q \) stored in a capacitor is directly proportional to the capacitance \( C \) and the voltage \( V \) across it, expressed by the formula:\[Q = C \times V\]This means that if the voltage applied to a capacitor remains constant, as the capacitance increases, more charge can be stored on the plates. Conversely, if the capacitance decreases, the amount of stored charge decreases, provided the voltage remains unchanged.
Capacitors are widely used in circuits for storing and releasing energy, filtering signals, and tuning radios, to mention a few applications. Understanding the relationship between capacitance and voltage is key to utilizing capacitors effectively in electronic devices.
  • Higher voltage across the capacitor increases the stored charge, given a constant capacitance.
  • If either the capacitance or the voltage is reduced to half, the stored charge will also drop by half.
Effect of Plate Separation on Capacitance
The separation between the plates of a capacitor is a crucial factor affecting its capacitance. The capacitance of a parallel-plate capacitor is inversely proportional to the distance \( d \) between the plates:\[C \propto \frac{1}{d}\]This means that as the distance between the plates increases, the capacitance decreases. Therefore, when the separation of the plates doubles, the capacitance is halved.
If a capacitor is connected to a constant voltage source, such as a battery, the decrease in capacitance leads to a proportional decrease in the charge stored on the plates. Consequently, in our example, doubling the plate separation while the capacitor remains connected to the 12.0 V battery results in halving the stored charge on the plates. This behavior is a reflection of the electric field distribution between the plates, where wider separation results in a decreased ability to store charge.
  • Increased plate separation results in reduced capacitance and stored charge for a fixed voltage.
  • The relationship is significant in applications requiring precise control of capacitance, such as tuning circuits.
Effect of Plate Area on Capacitance
The area \( A \) of the plates of a capacitor directly correlates with its capacitance. Specifically, capacitance is directly proportional to the plate area:\[C \propto A\]When the area increases, the capacitance of the capacitor also increases. This implies that a larger plate area allows more charge to be stored at the same voltage. In our problem's scenario, doubling the radius of the circular plates increases their area by a factor of four, since the area of a circle is calculated using \( A = \pi r^2 \).
With the capacitance increased fourfold and the battery supplying a constant voltage, the stored charge increases proportionally, four times in this example. This principle is particularly useful in designing capacitors where maximizing stored charge is essential, such as in energy storage devices.
  • Larger plate areas result in higher capacitance and increased charge storage capacity at the same voltage.
  • Doubling the radius results in a fourfold increase in plate area, significantly boosting charge capacity.

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Most popular questions from this chapter

BIO The electric egg. The eggs of many species undergo a rapid change in the electrical potential difference across the outer membrane when they are fertilized. This change in potential difference affects the physiological development of the eggs. The poterntial difference across the membrane is called the membrane potential, \(V_{m},\) defined as the inside potential minus the outside potential. The membrane potential \(V_{m}\) arises when protein enzymes use the energy available in ATP to actively expel sodium ions (Na') and accumulate potassium ions \(\left(\mathrm{K}^{+}\right) .\) Because the membrane of the unfertilized egg is selectively permeable to \(\mathrm{K}^{+},\) the \(V_{m}\) of the resting sea urchin egg is about \(-70 \mathrm{mV}\) ; that is, the inside has a potential of 70 \(\mathrm{mV}\) less than that of the outside. The egg membrane behaves as a capacitor with a specific capacitance of about 1\(\mu \mathrm{F} / \mathrm{cm}^{2} .\) When a sea urchin egg is fertilized, Na' channels in the membrane are opened, \(\mathrm{Na}^{+}\) enters the egg, and \(V_{m}\) rapidly changes to \(+30 \mathrm{mV},\) where it remains for several minutes. The concentration of \(\mathrm{Na}^{+}\) in the egg's interior is about 30 mmoles/liter (30 \(\mathrm{mM} )\) and 450 \(\mathrm{mM}\) in the surrounding sea water. The inside \(\mathrm{K}^{*}\) concentration is about 200 \(\mathrm{mM}\) and the outside \(\mathrm{K}^{+}\) is 10 \(\mathrm{mM} .\) A useful constant that connects electrical and chemical units is the Faraday number, which has a value of approximately \(10^{5}\) coulomb/mole. That is, an Avogadro number (a mole) of monovalent ions such as Na^ + or \(\mathrm{K}^{+}\) carries a charge of \(10^{5} \mathrm{C}\) . How many moles of \(\mathrm{Na}^{+}\) must move per unit area of membrane to change \(V_{m}\) from \(-70 \mathrm{mV}\) to \(+30 \mathrm{mV},\) making the assumption that the membrane behaves purely as a capacitor? A. \(10^{-4}\) mole \(/ \mathrm{cm}^{2}\) B. \(10^{-9}\) mole/cm \(^{2}\) C. \(10^{-12} \mathrm{mole} / \mathrm{cm}^{2}\) D. \(10^{-14} \mathrm{mole} / \mathrm{cm}^{2}\)

The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is \(2.50,\) and its dielectric strength is 50.0 \(\mathrm{MV} / \mathrm{m}\) . Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a 0.200\(\mu F\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Two oppositely charged identical insulating spheres, each 50.0 \(\mathrm{cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed 1.00 \(\mathrm{m}\) apart center to center Fig. 18.53 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b)\) on their surfaces, what will it read? (b) Which point, \(a\) or \(b,\) is at the higher potential? How can you know this without any calculations?

\(\bullet\) The plates of a parallel-plate capacitor are 3.28 \(\mathrm{mm}\) apart, and each has an area of 12.2 \(\mathrm{cm}^{2} .\) Each plate carries a charge of magnitude \(4.35 \times 10^{-8} \mathrm{C}\) . The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference between the plates? (c) What is the magnitude of the electric field between the plates?

A point charge \(q_{1}=+2.40 \mu C\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu C\) moves from the point \(x=0.150 \mathrm{m}, y=0,\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric forceon \(q_{2} ?\)
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