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Chapter 24: Problem 5

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) 120.0 \mu \mathrm{C}, (b) 60.0 \mu \mathrm{C}, (c) 480.0 \mu \mathrm{C}

Step by step solution

01

Calculate the Charge on Each Plate

From the given capacitor's capacitance (C) \(10.0 \mu \mathrm{F}\) and the battery voltage (V) \(12.0 \mathrm{~V}\), calculate the quantity of charge (Q) on each plate by utilizing the formula \( Q = C * V = 10.0 \mu \mathrm{F} * 12.0 \mathrm{~V} = 120.0 \mu \mathrm{C} \). Thus, the charge on each plate is 120.0 \mu \mathrm{C}.
02

Calculate the Charge after Doubling the Separation

When the separation is doubled while the capacitor remains connected to the battery, the voltage remains the same but the capacitance decreases by a factor of 2 (since capacitance is inversely related to the separation). Therefore, the new charge on each plate can be calculated as \( Q = (C/2) * V = (10.0 \mu \mathrm{F} / 2) * 12.0 \mathrm{~V} = 60.0 \mu \mathrm{C} \). Thus, if the separation doubles, the charge on each plate becomes 60.0 \mu \mathrm{C}.
03

Calculate the Charge after Doubling the Radius

If the radius of each plate is doubled, the area (and hence the capacitance) is quadrupled (since the area of a circle is given by \( \pi * (radius)^2 \)). The voltage remains the same since the capacitor is still connected to the battery. Therefore, the new charge on each plate can be calculated as \( Q = (4*C) * V = 4 * 10.0 \mu \mathrm{F} * 12.0 \mathrm{~V} = 480.0 \mu \mathrm{C} \). Thus, if the radius of the plates is doubled, the charge on each plate becomes 480.0 \mu \mathrm{C}.

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Key Concepts

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

Capacitance and Voltage Relationship
The relationship between capacitance and voltage is a fundamental concept in understanding how capacitors work. A capacitor stores electrical energy by holding a charge on its plates, and the ability to store this charge is defined by its capacitance. Capacitance is measured in farads and depends primarily on the physical characteristics such as the size of the plates and their distance apart.
To find out how much charge a capacitor can hold, we use the formula:
  • \( Q = C \times V \)
where \( Q \) is the charge, \( C \) is the capacitance, and \( V \) is the voltage applied.
This means that, for a given capacitance, the charge stored is directly proportional to the voltage applied across it. Increasing the voltage increases the amount of stored charge, as long as the capacitance remains unchanged. In practical applications, this relationship helps us determine how capacitors will behave in circuits at different voltage levels.
Effect of Plate Separation on Capacitance
The separation between the plates of a capacitor has a significant impact on its capacitance. Capacitors consist of two conductive plates separated by an insulating material, called a dielectric. The capacitance (
  • \( C = \frac{\varepsilon A}{d} \)
where \( \varepsilon \) is the permittivity of the dielectric, \( A \) is the area of the plates, and \( d \) is the separation between them) is inversely proportional to the distance \( d \) separating the plates.
Thus, if the distance between the plates is increased, the capacitance decreases. This reduction in capacitance means that the amount of charge that can be stored on the plates also decreases for a constant voltage, as indicated by the formula:\(
  • Q = \frac{C}{2} \times V \).

Understanding this concept is crucial for designing capacitors to fit specific electrical needs, ensuring devices operate efficiently without overloading the circuit.
Influence of Plate Area on Capacitance
The area of the plates in a capacitor plays a crucial role in determining the capacitance. Larger plate areas allow for more charge to be stored, which is reflected in the formula for capacitance, which depends directly on the plate area:
  • \( C = \frac{\varepsilon A}{d} \).
The larger the area of the plates of a capacitor, the greater its capacitance. This means that by increasing the plate area, we can enhance the charge storage ability, even if the voltage stays the same.
For example, as illustrated in the exercise, when the radius of the plates doubles, the area becomes four times larger (since the area of a circle is \( \pi \times \text{radius}^2 \)), and consequently, the capacitance increases fourfold.
This relationship helps engineers and technicians design capacitors that meet the requirements of various applications, ensuring the correct amount of charge storage capability for specific tasks.

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Most popular questions from this chapter

Two air-filled parallel-plate capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in parallel to a battery that has a voltage of \(36.0 \mathrm{~V}\) \(C_{1}=4.00 \mu \mathrm{F}\) and \(C_{2}=6.00 \mu \mathrm{F}\). (a) What is the total positive charge stored in the two capacitors? (b) While the capacitors remain connected to the battery, a dielectric with dielectric constant 5.00 is inserted between the plates of capacitor \(C_{1}\), completely filling the space between them. Then what is the total positive charge stored on the two capacitors? Does the insertion of the dielectric cause total charge stored to increase or decrease?

A parallel-plate capacitor has capacitance \(C_{0}=8.00 \mathrm{pF}\) when there is air between the plates. The separation between the plates is \(1.50 \mathrm{~mm}\). (a) What is the maximum magnitude of charge \(Q\) that can be placed on each plate if the electric field in the region between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{~V} / \mathrm{m} ?\) (b) A dielectric with \(K=2.70\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{~V} / \mathrm{m} ?\)

BIO Cell Membranes. Cell membranes (the walled enclosure around a cell) are typically about \(7.5 \mathrm{nm}\) thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about \(10 .\) (See Fig. \(\mathbf{P 2 4 . 4 8}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of \(85 \mathrm{mV}\) across its membrane. What is the electric field inside this membrane?

\- Polystyrene has dielectric constant 2.6 and dielectric strength \(2.0 \times 10^{7} \mathrm{~V} / \mathrm{m} .\) A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is \(80 \%\) of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage \(500.0 \mathrm{~V},\) the electric field between the plates is \(80 \%\) of the dielectric strength. What is the area of each plate if the capacitor stores \(0.200 \mathrm{~mJ}\) of energy under these conditions?

After combing your hair on a dry day, some of your hair stands up, forced away from your head by electrostatic repulsion. (a) Estimate the length \(L\) of your hair. (b) Using the average linear mass density of hair, which is \(65 \mu \mathrm{g} / \mathrm{cm},\) estimate the mass \(m\) of one of your hairs. (c) Estimate the number \(N\) of hairs that stand after combing. (d) Assume that the comb has taken away a charge \(-2 Q,\) and that your hair has therefore gained an amount of charge \(2 Q .\) Assume further that half of this charge resides next to your head and the other half is distributed at the ends of the \(N\) strands that stand up. Assume the electrostatic force that lifted a hair was twice its weight. Show that this leads to \(2 m g=\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2} / N}{L^{2}}\) Use this equation to estimate the charge \(Q\) that resides on your head. (e) If your head were a sphere with radius \(R\), it would have a capacitance of \(4 \pi \epsilon_{0} R .\) Estimate the radius of your head; then use your estimate to determine your head's capacitance. (f) Use your result to estimate the potential attained due to combing. (The surprising result illustrates an interesting feature of static electricity.)
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