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24.1. A capacitor has a capacitance of 7.28\(\mu \mathrm{F}\) . What amount of charge must be placed on each of its plates to make the potential difference between its plates equal to 25.0 \(\mathrm{V} ?\)

Short Answer

Expert verified
The charge on each plate is 182 μC.

Step by step solution

01

Identify Given Values

We are given the capacitance of the capacitor as \( C = 7.28 \mu \text{F} \) and the potential difference \( V = 25.0 \text{V} \). Convert the capacitance to farads: \( 7.28 \mu \text{F} = 7.28 \times 10^{-6} \text{F} \).
02

Apply the Capacitance Formula

The relationship between charge \( Q \), capacitance \( C \), and potential difference \( V \) is given by the formula: \( Q = C \times V \).
03

Substitute Values

Substitute the given values into the formula: \( Q = 7.28 \times 10^{-6} \text{F} \times 25.0 \text{V} \).
04

Calculate the Charge

Perform the calculation:\[Q = 7.28 \times 10^{-6} \times 25.0 = 1.82 \times 10^{-4} \text{C} = 182 \mu \text{C}.\]
05

Verify Units and Result

Ensure that the units are consistent and reasonable. The calculated charge is \( 182 \mu \text{C} \), which fits given that both the capacitance and potential difference are typical values.

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Key Concepts

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

Charge on Capacitor Plates
When a capacitor is charged, an equal amount of positive and negative charge accumulates on each of its plates. This charge is denoted by the symbol \(Q\). Essentially, a capacitor stores energy in the form of electrical charge. The charge is directly related to the capacitor's ability to hold this charge and the potential difference across its plates.
  • To determine the charge on the plates, we rely on the relationship between charge \(Q\), capacitance \(C\), and potential difference \(V\).
  • The charge on the plates is measured in coulombs, a unit that represents electric charge.
  • In practical terms, this stored charge can be discharged later to power electrical devices.
Understanding how charge behaves in a capacitor helps us comprehend the basic principles of electricity and its applications in circuits. In our example, you discovered that the charge needed to reach a potential difference of 25.0 volts with a capacitance of 7.28\(\mu\mathrm{F}\) is 182\(\mu\mathrm{C}\).
Capacitance Formula
The capacitance formula is a fundamental equation used to relate three important components: charge \(Q\), capacitance \(C\), and potential difference \(V\). According to the formula:\[ Q = C \times V \]
  • \(Q\) represents the charge stored in the capacitor measured in coulombs.
  • \(C\) stands for capacitance, the ability of the capacitor to store charge, measured in farads (F).
  • \(V\) is the potential difference across the capacitor's plates, measured in volts (V).
This equation allows you to calculate the amount of charge that develops on the capacitor's plates when the capacitance and potential difference are known. It's particularly useful in studying circuit behaviors as it can predict how much energy a capacitor can store or release in a circuit.
Potential Difference
The potential difference, also known as voltage, is the measure of work needed to move a unit charge from one point to another in an electric field. In the context of a capacitor, it refers to the voltage across the capacitor's plates. When we say a capacitor has a potential difference, we're talking about the energy per unit charge that's available to move charges between the plates.
  • Potential difference is measured in volts (V).
  • A higher potential difference implies more energy is stored in the capacitor.
  • It determines, along with the capacitance, the total charge stored on the capacitor's plates.
A crucial aspect of potential difference in capacitors is how it becomes the deciding factor in determining both the charge and energy storage capacity of the capacitor. In our example, with a potential difference of 25.0 volts, it directly affected the amount of charge that was calculated to be 182\(\mu\mathrm{C}\). This emphasizes the role voltage plays in the functionality and overall energy storage of capacitors.

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Most popular questions from this chapter

24.46. 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 \(\mathrm{pC}\) 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 insered? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

24.20. Two parallel plate vacuum capacitors have plate spacings \(d_{1}\) and \(d_{2}\) and equal plate areas \(A\) . Show that when the capacitors are connected in series, the eqnivalent capacitance is the same as for a single capacitor with plate area \(A\) and spacing \(d_{1}+d_{2}\) .

24.13. A spherical capacitor contains a charge of 3.30 \(\mathrm{nC}\) when connected to a potential difference of 220 \(\mathrm{V}\) . If its plates are sepa- rated by vacuum and the inner radius of the outer shell is \(4.00 \mathrm{cm},\) calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

24.12. A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 \(\mathrm{cm}\) and the capacitance is 116 \(\mathrm{pF}\) . (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 \(\mathrm{V}\) , what is the magnitude of charge on each sphere?

24.21. Two parallel-plate vacuum capacitors have areas \(A_{1}\) and \(A_{2}\) and equal plate spacings \(d .\) Show that when the capacitors are connected in parallel, the equivalent capacitance is the same as for a single capacitor with plate area \(A_{1}+A_{2}\) and spacing \(d\) .

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