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A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

Short Answer

Expert verified
(a) Initial energy: 3.60 mJ; After dielectric: 13.5 mJ. (b) Energy increased by 9.9 mJ.

Step by step solution

01

Calculate the initial energy stored in the capacitor

The energy stored in a capacitor is given by the formula: \[ U = \frac{1}{2}CV^2 \]. We first calculate the energy stored in the capacitor before the dielectric is added. Here, \( C = 12.5 \mu F = 12.5 \times 10^{-6} F \) and \( V = 24.0 \) V. Substituting these values into the formula gives: \[ U = \frac{1}{2} \times 12.5 \times 10^{-6} \times (24.0)^2 \]. Evaluating this gives the initial energy.
02

Calculate the new capacitance with the dielectric

The capacitance with a dielectric is calculated using \( C' = K \cdot C \), where \( K \) is the dielectric constant. Inserting \( K = 3.75 \) and the original capacitance \( C = 12.5 \times 10^{-6} F \): \[ C' = 3.75 \times 12.5 \times 10^{-6} \]. This will give us the new capacitance with the dielectric.
03

Calculate the energy stored with the dielectric

Using the new capacitance found in Step 2 and the same voltage (since the voltage stays constant), calculate the new energy: \[ U' = \frac{1}{2}C'V^2 \]. Substitute \( C' \) from Step 2 and continue with \( V = 24.0 \) V to find the energy stored in the capacitor after the dielectric is inserted.
04

Calculate the change in energy

Subtract the initial energy stored from the new energy stored to find the change in energy: \[ \Delta U = U' - U \]. Use the values calculated in Steps 1 and 3 to determine \( \Delta U \).
05

Determine the nature of energy change

If \( \Delta U \) is positive, it means the energy increased after inserting the dielectric. If \( \Delta U \) is negative, the energy decreased. Compare \( U' \) with \( U \) to find out if the energy increased or decreased.

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Key Concepts

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

Dielectric Constant
A dielectric material is a substance that, when inserted between the plates of a capacitor, affects its electrical properties. The key purpose of a dielectric is to increase the capacitance. This change is measured by a dielectric constant, denoted as "\( K \)". The dielectric constant is a dimensionless number that indicates how much a dielectric material can increase the capacitance compared to a vacuum.
In simple terms:
  • "Dielectric Constant (\( K \)) = \text{Capacitance with dielectric} / \text{Capacitance without dielectric}"
  • The higher the dielectric constant, the greater the increase in capacitance.
  • A vacuum has a dielectric constant of 1, while materials like paper or porcelain have higher values due to their ability to store more electrical energy.
By placing a dielectric between capacitor plates, you effectively enhance stored energy without changing the voltage. In our problem, the dielectric constant is 3.75, signaling that the capacitance increases by a factor of 3.75 when the dielectric is fully applied.
Capacitance
Capacitance, represented by "\( C \)", is the ability of a capacitor to store charge per unit voltage. It depends on the physical attributes of the capacitor, like the area of the plates, the distance between them, and any dielectric material placed between them.
Key points to understand capacitance are:
  • Measured in Farads (\( F \)), which quantify the charge stored per volt.
  • Larger plate areas or reduced distance between plates generally lead to higher capacitance.
  • Introducing a dielectric material increases capacitance by reducing the electric field strength between the plates.
In the given exercise, the initial capacitance without the dielectric is 12.5 \( \mu \)F. With a dielectric constant of 3.75 fully introduced, the capacitance increases, showcasing how effective a dielectric can be at boosting energy storage capabilities.
Potential Difference
The potential difference, indicated by "\( V \)", is the voltage across the capacitor's plates. It signifies the energy potential per charge in an electric field and drives the movement of electric charge between the plates.Essential elements to grasp potential difference:
  • Measured in Volts (\( V \)). It communicates the pressure from an electric circuit's power source that pushes charged electrons (current) through a conducting loop.
  • In capacitors, charge collects on plates of opposite potentials, creating a potential difference.
  • In this specific exercise, the potential difference remains constant at 24 V even after the dielectric insertion, showing that energy change occurs due to capacitance alteration, not a difference in voltage.
Maintaining a constant potential difference highlights the role of the dielectric and capacitance in energy storage respective changes.
Energy Stored in Capacitors
Energy storage in capacitors involves the electric potential between the capacitor's plates, where energy is stored as an electric field. This stored energy (\( U \)) is crucial for devices requiring bursts of power.The formula for energy stored in a capacitor is: \[ U = \frac{1}{2}CV^2 \]Key insights on capacitor energy storage:
  • It relies on both the capacitance \( C \) and the potential difference \( V \).
  • Adding a dielectric increases the capacitance \( C \), enhancing energy storage \( U \) as the relationship is quadratic with voltage.
  • In our scenario, energy increases after dielectric insertion due to increased capacitance, calculated using both pre and post-insertion capacitance values.
This exercise illustrates how dielectric materials can influence a capacitor's ability to store energy by effectively adjusting the stored energy equation through changes in capacitance.

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Most popular questions from this chapter

An air capacitor is made by using two flat plates, each with area A, separated by a distance \(d\). Then a metal slab having thickness \(a\) (less than \(d\)) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (\(\textbf{Fig. P24.62}\)). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C\(_0\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a\) \(\rightarrow\) 0 and \(a\) \(\rightarrow\) d.

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10. (See \(\textbf{Fig. P24.48}\).) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 \(\times\) 10\(^2\) V. The electric field between the plates is to be no greater than 1.00 \(\times\) 10\(^4\) N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

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