Question

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

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

Step-by-step solution

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.

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.

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.

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 \( \Delta U \).

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.

Key concepts

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.