Question

The capacitance of a parallel plate air capacitor is \(1 \mu \mathrm{F}\). This region is now filled with a dielectric slab of dielectric constant 4 and capacitor is connected with battery of EMF 5 volt and zero internal resistance. Now slab is taken out, then during the removal of slab then which of the following options is incorrect: (1) \(-15 \mu \mathrm{C}\) charge flows through the cell (2) \(75 \mu\) energy is absorbed by the cell (3) energy stored in the capacitor is reduced by \(75 \mu \mathrm{J}\) (4) the external agent has to do \(37.5 \mu \mathrm{J}\) amount of work to take out the slab (5) charge on capacitor reduced by \(15 \mu \mathrm{C}\)

Short answer

Option 2 is incorrect: 75 μJ energy absorption should be 37.5 µJ.

Step-by-step solution

Identify given data and initial conditions

The capacitance of the capacitor with air: \(C = 1 \, \text{μF}\)Dielectric constant: \(k = 4\)Battery EMF: \(V = 5 \, \text{V}\)

Calculate capacitance with dielectric

When the dielectric is inserted, the capacitance becomes \(C' = kC = 4 \times 1 \, \text{μF} = 4 \, \text{μF}\).

Determine initial charge with dielectric

Using the formula \(Q' = C'V\): \(Q' = 4 \, \text{μF} \times 5 \, \text{V} = 20 \, \text{μC}\).

Determine initial energy stored with dielectric

Using the formula \(U' = \frac{1}{2} C' V^2\): \[U' = \frac{1}{2} \times 4 \, \text{μF} \times (5 \, \text{V})^2 = 50 \, \text{μJ}\].

Remove the dielectric and find new capacitance

After removal of the dielectric, the capacitance returns to \(C = 1 \, \text{μF}\).

Find final charge after dielectric removal

Using the charge formula for the new capacitance: \(Q = CV\)\(Q = 1 \, \text{μF} \times 5 \, \text{V} = 5 \, \text{μC}\).

Calculate the change in charge

The change in charge: \(\text{ΔQ} = Q' - Q = 20 \, \text{μC} - 5 \, \text{μC} = 15 \, \text{μC}\).Therefore, charge reduced by the capacitor is \(15 \, \text{μC}\).

Determine the charge flow through the battery

Since the correct charge change happens through the battery, \(15 \, \text{μC}\) will flow through the battery.

Calculate the energy stored after removing the dielectric

Using \(U = \frac{1}{2} C V^2\):\[U = \frac{1}{2} \times 1 \, \text{μF} \times (5 \, \text{V})^2 = 12.5 \, \text{μJ}\].

Compute the change in stored energy

The change in the energy stored:\[ΔU = U' - U = 50 \, \text{μJ} - 12.5 \, \text{μJ} = 37.5 \, \text{μJ}\].

Evaluate the work done by the external agent

Work done by the external agent is same as the energy change:\(W = 37.5 \, \text{μJ}\).

Determine incorrect option

Option 2 states that 75 μJ energy is absorbed by the cell, but from calculations:Energy In = 37.5 μJ < Energy Out =75μJSo option 2 is incorrect.

Key concepts

Dielectric Constant

The dielectric constant, also known as the relative permittivity, is a measure of how much an insulating material can increase the capacitance of a capacitor. In simple terms, it quantifies how well a material can store electrical energy in an electric field compared to vacuum. When a dielectric material is inserted between the plates of a capacitor, the capacitance increases because the dielectric material reduces the electric field strength inside the capacitor, allowing more charge to be stored for the same applied voltage.

In the given exercise, the dielectric constant is 4. This means that when the dielectric slab is inserted into the initially air-filled capacitor, the capacitance becomes four times larger. This can be represented by the formula:

\[ C' = kC \]
where:

• \( C' \) is the capacitance with the dielectric.
• \( k \) is the dielectric constant.
• \( C \) is the initial capacitance without the dielectric.

In our case:

\[ C' = 4 \times 1 \text{ μF} = 4 \text{ μF} \]

Capacitance

Capacitance is a measure of a capacitor's ability to store electrical charge. It is defined as the ratio of the electric charge (Q) stored on the capacitor to the voltage (V) across its plates:

\[ C = \frac{Q}{V} \]
In simpler terms, it tells us how much charge the capacitor can hold when a specific voltage is applied across its plates. The unit of capacitance is the Farad (F), which is often expressed in microfarads (μF) for practical applications.

For a parallel plate capacitor without a dielectric, the capacitance is given. In our exercise, the initial capacitance is:

\[ C = 1 \text{ μF} \]
When the dielectric slab with a dielectric constant of 4 is inserted, the capacitance increases to:

\[ C' = 4 \text{ μF} \]
To further understand, let's see how the charge stored changes. Initially, with the dielectric inserted:

\[ Q' = C' \times V = 4 \times 5 = 20 \text{ μC} \]
After removing the dielectric:

\[ Q = C \times V = 1 \times 5 = 5 \text{ μC} \]
This reduction in charge affects the energy stored and the work required to remove the dielectric, as we will explore next.

Stored Energy

The energy stored in a capacitor is the energy contained in the electric field between its plates. This stored energy can be calculated with the formula:

\[ U = \frac{1}{2} C V^2 \]
In simple terms, this formula tells us that the energy stored is dependent on the capacitance and the square of the voltage. For the given exercise, we have two key scenarios: before and after the dielectric is removed.

Initially, with the dielectric slab inserted:

\[ U' = \frac{1}{2} \times 4 \text{ μF} \times (5 \text{ V})^2 = 50 \text{ μJ} \]
After removing the dielectric, the capacitor returns to its initial capacitance:

\[ U = \frac{1}{2} \times 1 \text{ μF} \times (5 \text{ V})^2 = 12.5 \text{ μJ} \]
The change in the stored energy is:

\[ ΔU = U' - U = 50 \text{ μJ} - 12.5 \text{ μJ} = 37.5 \text{ μJ} \]
The removal of the dielectric slab thus results in a significant reduction in the stored energy. This change in energy implies that work has been done to remove the slab, supporting the calculated value in the step-by-step solution. However, it's important to note that, contrary to the incorrect option 2 in the problem, only 37.5 μJ is absorbed by the cell.