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Chapter 18: Problem 52

A 4.00\(\mu \mathrm{F}\) and a 6.00\(\mu \mathrm{F}\) capacitor are connected in series, and this combination is connected across a 48.0 \(\mathrm{V}\) potential difference. Calculate (a) the charge on each capacitor and (b) the potential difference across each of them.

Short Answer

Expert verified
Each capacitor has a charge of 115.2 渭C. The potential differences are 28.8 V and 19.2 V across the 4.00 渭F and 6.00 渭F capacitors, respectively.

Step by step solution

01

Calculate the equivalent capacitance

When capacitors are connected in series, the equivalent capacitance \( C_{eq} \) can be calculated using the formula: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]Substituting the given values, \( C_1 = 4.00 \, \mu\mathrm{F} \) and \( C_2 = 6.00 \, \mu\mathrm{F} \): \[ \frac{1}{C_{eq}} = \frac{1}{4.00} + \frac{1}{6.00} \]\[ \frac{1}{C_{eq}} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \]\[ C_{eq} = \frac{12}{5} = 2.40 \, \mu\mathrm{F} \]
02

Calculate the charge on the equivalent capacitor

To find the charge \( Q \) on the series combination, use the formula:\[ Q = C_{eq} \cdot V \]where \( V = 48.0 \, \mathrm{V} \). Substituting the equivalent capacitance:\[ Q = 2.40 \, \mu\mathrm{F} \times 48.0 \, \mathrm{V} \]\[ Q = 115.2 \, \mu\mathrm{C} \]The charge on each capacitor in series is the same, so each capacitor carries a charge of \( 115.2 \, \mu\mathrm{C} \).
03

Calculate the potential difference across each capacitor

Using the charge calculated in Step 2, find the potential difference \( V_1 \) across the 4.00 \( \mu\mathrm{F} \) capacitor:\[ V_1 = \frac{Q}{C_1} = \frac{115.2 \, \mu\mathrm{C}}{4.00 \, \mu\mathrm{F}} = 28.8 \, \mathrm{V} \]Then, find the potential difference \( V_2 \) across the 6.00 \( \mu\mathrm{F} \) capacitor:\[ V_2 = \frac{Q}{C_2} = \frac{115.2 \, \mu\mathrm{C}}{6.00 \, \mu\mathrm{F}} = 19.2 \, \mathrm{V} \]

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Key Concepts

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

Equivalent Capacitance
When capacitors are connected in series, the overall capacitance behaves differently compared to when they're connected in parallel. To find the equivalent capacitance \( C_{eq} \) for capacitors in series, you use the reciprocal formula:
  • \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \)
This formula means that the total capacitance is less than the smallest capacitance in the series. In our example, with a 4.00 \(\mu\mathrm{F}\) and a 6.00 \(\mu\mathrm{F}\) capacitor connected in series:
  • The calculation becomes \( \frac{1}{C_{eq}} = \frac{1}{4.00\,\mu\mathrm{F}} + \frac{1}{6.00\,\mu\mathrm{F}} \)
  • After combining the fractions, the equivalent capacitance comes out to be \(2.40\,\mu\mathrm{F}\).
This reduced equivalent capacitance occurs because, in a series circuit, each capacitor shares the same charge, leading to an eased overall effect on storing energy compared to capacitors connected in parallel.
Charge on Capacitors
A key aspect of capacitors in series is that they all share the same charge. This is because there is only one path for the charge to flow; hence, the charge on each capacitor is identical.
  • To calculate the charge \( Q \) on each capacitor, use the formula: \( Q = C_{eq} \cdot V \)
  • Where \( C_{eq} \) is the equivalent capacitance and \( V \) is the applied voltage across the series.
In the given problem, the equivalent capacitance is \(2.40\,\mu\mathrm{F}\) and the voltage is \(48.0\,\mathrm{V}\), leading to:
  • \( Q = 2.40\,\mu\mathrm{F} \times 48.0\,\mathrm{V} = 115.2\,\mu\mathrm{C} \)
So, each of the capacitors in the series has a charge of \(115.2\,\mu\mathrm{C}\) on it. This uniformity of charge simplifies the analysis and behavior prediction of series circuits.
Potential Difference Across Capacitors
When dealing with series capacitors, each capacitor can have a different potential difference across it, even though they share the same charge. To find the potential difference across each capacitor, you use:
  • \( V = \frac{Q}{C} \)
For our specific capacitors:
  • For the 4.00 \(\mu\mathrm{F}\) capacitor: \( V_1 = \frac{115.2\,\mu\mathrm{C}}{4.00\,\mu\mathrm{F}} = 28.8\,\mathrm{V} \)
  • For the 6.00 \(\mu\mathrm{F}\) capacitor: \( V_2 = \frac{115.2\,\mu\mathrm{C}}{6.00\,\mu\mathrm{F}} = 19.2\,\mathrm{V} \)
Although the total voltage across the series is \(48.0\,\mathrm{V}\), it's distributed differently across each capacitor. The division depends on individual capacitances, with larger capacitances receiving a smaller share of the total voltage. Understanding this distribution is crucial for ensuring each capacitor's voltage ratings are not exceeded in a circuit design.

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Most popular questions from this chapter

Two very large charged parallel metal plates are 10.0 \(\mathrm{cm}\) apart and produce a uniform electric field of \(2.80 \times 10^{6} \mathrm{N} / \mathrm{C}\) between them. A proton is fired perpendicular to these plates with an initial speed of 5.20 \(\mathrm{km} / \mathrm{s}\) , starting at the middle of the negative plate and going toward the positive plate. How much work has the electric field done on this proton by the time it reaches the positive plate?

(a) You find that if you place charges of \(\pm 1.25 \mu \mathrm{C}\) on two separated metal objects, the potential difference between them is 11.3 \(\mathrm{V}\) . What is their capacitance? (b) A capacitor has a capacitance of 7.28\(\mu \mathrm{F}\) . What amount of excess charge must be placed on each of its plates to make the potential difference between the plates equal to 25.0 \(\mathrm{V}\) ?

When two point charges are a distance \(R\) apart, their potential energy is \(-2.0 \mathrm{J} .\) How far far (in terms of \(R )\) should they be from each other so that their potential energy is \(-6.0 \mathrm{J} ?\)

A 5.80\(\mu\) F parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3} .\)

A gold nucleus has a radius of \(7.3 \times 10^{-15} \mathrm{m}\) and a charge of \(+79 e .\) Through what voltage must an \(\alpha\) -particle, with its charge of \(+2 e,\) be accelerated so that it has just enough energy to reach a distance of \(2.0 \times 10^{-14} \mathrm{m}\) from the surface of a gold nucleus? (Assume the gold nucleus remains stationary and can be treated as a point charge.)
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