Ch 24: Capacitance and Dielectrics

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 24: Capacitance and Dielectrics

Problem 20

Chapter 24, Problem 20

In Fig. E $24.20$ , $C sub 1 equals 6.00$ $μ$ F, $C sub 2 equals 3.00$ $μ$ F, and $C sub 3 equals 5.00$ $μ$ F. The capacitor network is connected to an applied potential $V_{a b}$ .
(a) After the charges on the capacitors have reached their final values, the charge on $C sub 2$ is $30.0$ mC. What are the charges on capacitors $C sub 1$ and $C sub 3$ ?
(b) What is the applied voltage $V_{a b}$ ?

Verified step by step guidance

First, identify the configuration of the capacitors. In the given circuit, C1 and C2 are in parallel, and their combination is in series with C3.

Calculate the equivalent capacitance of C1 and C2 in parallel using the formula:

C_{eq} = C_{1} + C_{2}

. Substitute the given values:

C_{eq} = 6.00 + 3.00

Next, calculate the total equivalent capacitance of the circuit, which is the series combination of the parallel equivalent capacitance and C3. Use the formula for capacitors in series:

C_{total} = \frac{1}{\frac{1}{C_{eq}}}

To find the charge on C1 and C3, use the fact that the charge on capacitors in series is the same. Since the charge on C2 is given as 30.0 mC, the charge on C1 and C3 will also be 30.0 mC.

Finally, calculate the applied voltage Vab using the formula:

V = \frac{Q}{C}

. For the series combination, use the total charge and total equivalent capacitance:

V = \frac{30.0}{C_{total}}

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage. It is defined as C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the capacitor. In this problem, the capacitance values of C1, C2, and C3 are given, which will help determine the charge on each capacitor when connected in a circuit.

Series and Parallel Capacitors

Capacitors can be arranged in series or parallel configurations, affecting the total capacitance of the circuit. In series, the total capacitance decreases, while in parallel, it increases. The arrangement of C1, C2, and C3 in the circuit diagram influences how the charges distribute among them and how to calculate the total voltage across the network.

Charge Conservation

Charge conservation states that the total charge in a closed system remains constant. In this circuit, the charge on C2 is given, and using charge conservation principles, we can find the charges on C1 and C3. This concept is crucial for solving part (a) of the question, as it allows us to relate the charges on the capacitors based on their configuration.

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Textbook Question

For the system of capacitors shown in Fig. E $24.16$ , find the equivalent capacitance between $b$ and $c$ .

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Textbook Question

Figure E $24.14$ shows a system of four capacitors, where the potential difference across ab is $50.0$ V. How much charge is stored in each of the $10.0$ - $μ$ F and the $9.0$ - $μ$ F capacitors?

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Textbook Question

A $5.80$ - $μ$ F, parallel-plate, air capacitor has a plate separation of $5.00$ mm and is charged to a potential difference of $400$ V. Calculate the energy density in the region between the plates, in units of J/m³.

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Figure E $24.14$ shows a system of four capacitors, where the potential difference across ab is $50.0$ V. How much charge is stored by this combination of capacitors?

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Textbook Question

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 uC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

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Textbook Question

An air capacitor is made from two flat parallel plates $1.50$ mm apart. The magnitude of charge on each plate is $0.0180$ $μ$ C when the potential difference is $200$ V.

(a) What is the capacitance?

(b) What is the area of each plate?

(c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of $3.0 \times 10^{6}$ V/m.)

(d) When the charge is $0.0180$ $μ$ C, what total energy is stored?