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Chapter 24: Problem 30

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?

Short Answer

Expert verified
The total positive charge initially stored in the two capacitors is 360 渭C. After a dielectric with a dielectric constant of 5 is inserted into capacitor C1, the total positive charge stored on the two capacitors increases to 936 渭C.

Step by step solution

01

STEP 1: Calculate the total initial charge on the capacitors

Capacitance (\(C\)) is defined as the ratio of the magnitude of the charge (\(Q\)) on each conductor to the magnitude of the potential difference (\(V\)) between the conductors. Therefore, the charge on each capacitor can be calculated using the formula: \(Q = CV\). For capacitor \(C_1\), the charge is \(Q_1 = C_1V = 4 \mu F * 36 V = 144 \mu C\); and for capacitor \(C_2\), the charge is \(Q_2 = C_2V = 6 \mu F * 36 V = 216 \mu C\). The total initial charge on the capacitors is then \(Q_{total} = Q_1 + Q_2 = 144 \mu C + 216 \mu C = 360 \mu C\).
02

STEP 2: Calculate the new capacitance after inserting the dielectric

When a dielectric of dielectric constant \(K\) is inserted between the plates of a capacitor, the capacitance increases by a factor of \(K\). Therefore, the new capacitance of capacitor \(C_1\) after inserting the dielectric is \(C'_1 = KC_1 = 5*4 \mu F = 20 \mu F\). The capacitance of capacitor \(C_2\) remains unchanged because it does not contain the dielectric.
03

STEP 3: Calculate the total charge on the capacitors after inserting the dielectric

Now we can calculate the new charge on each capacitor after inserting the dielectric. For capacitor \(C_1\), the new charge is \(Q'_1 = C'_1 V = 20 \mu F * 36 V = 720 \mu C\); and for capacitor \(C_2\), the charge remains \(Q_2 = 216 \mu C\) because its capacitance did not change. The total charge on the capacitors after inserting the dielectric is \(Q'_{total} = Q'_1 + Q_2 = 720 \mu C + 216 \mu C = 936 \mu C\). Since \(Q'_{total} > Q_{total}\), the total charge stored has increased as a result of inserting the dielectric.

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a simple device used to store electrical energy. It consists of two conductive plates separated by a small distance, typically creating a uniform electric field between them. These plates are each capable of holding an equal but opposite charge. The difference in charge gives rise to a voltage across the plates.
  • The capacitance of a parallel-plate capacitor depends on the surface area of the plates and the distance between them.
  • Larger plate area increases capacitance, while a greater distance between plates decreases it.
  • The relationship for capacitance in a parallel-plate setup is given by the formula: \[ C = \frac{\varepsilon_0 \cdot A}{d} \]where \(C\) is the capacitance, \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of one plate, and \(d\) is the separation between the plates.
Understanding this concept is crucial as it serves as the foundation for calculating charge storage in capacitors, both with and without dielectrics.
Dielectric Constant
The dielectric constant is a measure of a material's ability to increase the capacitance of a capacitor beyond just the basic air-filled scenario. In essence, it's a multiplier for capacitance when a particular dielectric material is placed between the plates.
  • Dielectrics are insulating materials, such as glass or plastic, that increase a capacitor鈥檚 ability to store charge.
  • The dielectric constant \(K\) is a dimensionless quantity, with air's dielectric constant as close to 1.
  • Inserting a dielectric increases the capacitance to \(C' = K \cdot C\).
For example, if the dielectric constant is 5, the new capacitance becomes five times the original. This principle is used in the exercise to calculate increased charge storage when a dielectric is inserted in capacitor \(C_1\).
Capacitance
Capacitance is a fundamental concept in electronics and physics. It measures how much charge a capacitor can store per unit voltage.
  • Represented as \(C\), the unit of capacitance is the farad (\(F\)).
  • The formula used to determine the charge \(Q\) stored in a capacitor is \[ Q = CV \] where \(C\) is the capacitance and \(V\) is the voltage.
  • A higher capacitance or voltage results in an increased ability to store charge.
In our exercise, the individual charges stored before and after inserting the dielectric were calculated using these principles, demonstrating how capacitance is central to understanding capacitor behavior.

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Most popular questions from this chapter

A parallel-plate air capacitor is made by using two plates \(12 \mathrm{~cm}\) 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 capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius \(12.5 \mathrm{~cm},\) and the outer sphere has radius \(14.8 \mathrm{~cm} .\) A potential difference of \(120 \mathrm{~V}\) is applied to the capacitor. (a) What is the energy density at \(r=12.6 \mathrm{~cm}\), just outside the inner sphere? (b) What is the energy density at \(r=14.7 \mathrm{~cm},\) just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

A parallel-plate air capacitor is to store charge of magnitude \(240.0 \mathrm{pC}\) on each plate when the potential difference between the plates is \(42.0 \mathrm{~V}\). (a) If the area of each plate is \(6.80 \mathrm{~cm}^{2}\), what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude \(240.0 \mathrm{pC}\) on each plate?

A \(5.00 \mathrm{pF},\) parallel-plate, air-filled capacitor with circular 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?

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 separated 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.
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