Ch 24: Capacitance and Dielectrics

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 24: Capacitance and Dielectrics

Problema 7

Capitolo 24, Problema 7

A parallel-plate air capacitor is to store charge of magnitude $240.0$ pC on each plate when the potential difference between the plates is $42.0$ V.
(a) If the area of each plate is $6.80$ cm², 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 $240.0$ pC on each plate?

Guida verificata passo dopo passo

Step 1: Begin by understanding the relationship between charge (Q), capacitance (C), and potential difference (V) in a capacitor, given by the formula:

Q = C V

. Here, Q is the charge, C is the capacitance, and V is the potential difference.

Step 2: For a parallel-plate capacitor, the capacitance can be calculated using the formula:

C = \frac{ε A}{d}

, where

ε

is the permittivity of free space (approximately

8.85 \times 10^- 12

F/m), A is the area of the plates, and d is the separation between the plates.

Step 3: Rearrange the formula from Step 2 to solve for the separation distance d:

d = \frac{ε A}{C}

. Substitute the values for

ε

, A (converted to m²), and C (calculated from Step 1 using

C = \frac{Q}{V}

) to find d.

Step 4: For part (b), if the separation between the plates is doubled, the new separation is

2 d

. Use the formula for capacitance again,

C = \frac{ε A}{2 d}

, to find the new capacitance.

Step 5: Finally, use the relationship

Q = C V

again to solve for the new potential difference V required to store the same charge Q with the new capacitance calculated in Step 4.

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Capacitance

Capacitance is the ability of a capacitor to store charge per unit voltage, defined by the formula C = Q/V, where C is capacitance, Q is charge, and V is voltage. For parallel-plate capacitors, capacitance is also given by C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between them.

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Percorso guidato

08:02

Capacitors & Capacitance (Intro)

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by an insulating material (dielectric). The capacitance depends on the area of the plates and the distance between them, as well as the dielectric material. It is a fundamental component in circuits for storing electrical energy and controlling voltage levels.

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Percorso guidato

08:53

Parallel Plate Capacitors

Permittivity of Free Space

The permittivity of free space, denoted as ε₀, is a constant that characterizes the ability of a vacuum to permit electric field lines. It is approximately 8.85 x 10⁻¹² F/m (farads per meter) and is crucial in calculating the capacitance of capacitors, especially in the formula C = ε₀A/d for parallel-plate capacitors.

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Percorso guidato

5:05

Spinning Space Station

Pratica correlata

Domanda del libro di testo

A $5.00$ - $μ$ F parallel-plate capacitor is connected to a $12.0$ V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates.

(a) A voltmeter is connected across the two plates without discharging them. What does it read?

(b) What would the voltmeter read if (i) the plate separation were doubled; (ii) the radius of each plate were doubled but their separation was unchanged?

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Domanda del libro di testo

The plates of a parallel-plate capacitor are $2.50$ mm apart, and each carries a charge of magnitude $80.0$ nC. The plates are in vacuum. The electric field between the plates has a magnitude of $4.00 \times 10^{6}$ V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

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A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is $10.0$ pC. The inner cylinder has radius $0.50$ mm, the outer one has radius $5.00$ mm, and the length of each cylinder is $18.0$ cm.

(a) What is the capacitance?

(b) What applied potential difference is necessary to produce these charges on the cylinders?

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Domanda del libro di testo

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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Domanda del libro di testo

A parallel-plate air capacitor of capacitance $245$ pF has a charge of magnitude $0.148$ $μ$ C on each plate. The plates are $0.328$ mm apart.

(a) What is the potential difference between the plates?

(b) What is the area of each plate?

(d) What is the surface charge density on each plate?

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Domanda del libro di testo

A spherical capacitor contains a charge of $3.30$ nC when connected to a potential difference of $220$ V. If its plates are separated by vacuum and the inner radius of the outer shell is $4.00$ 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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