Understanding Parallel Plate Capacitor Capacitance

The question asks how the capacitance of a parallel plate capacitor changes when both the distance between the plates is altered and a dielectric material is introduced into the space between them.

The capacitance \(C\) of a parallel plate capacitor is determined by the area of the plates \(A\), the distance between the plates \(d\), and the permittivity of the material between the plates \(\epsilon\). The formula is given by:

\(C = \frac{\epsilon A}{d}\)

The permittivity \(\epsilon\) of a material is related to the permittivity of free space \(\epsilon_0\) by the dielectric constant \(\kappa\), such that \(\epsilon = \kappa \epsilon_0\). For air, the dielectric constant is approximately \(\kappa_{air} \approx 1\).

Initial State Analysis: Capacitor with Air

Initially, the capacitor has air between the plates, and its capacitance is given as 6 pF. Let the initial distance between the plates be \(d_1\) and the area of the plates be \(A\).

The initial capacitance \(C_1\) is:

\(C_1 = \frac{\kappa_{air} \epsilon_0 A}{d_1}\)

Since \(\kappa_{air} \approx 1\), we have:

\(C_1 = \frac{\epsilon_0 A}{d_1} = 6 \text{ pF}\)

Final State Analysis: Capacitor with Dielectric and Reduced Distance

In the new scenario, the distance between the plates is reduced to half, so the new distance \(d_2\) is:

\(d_2 = \frac{d_1}{2}\)

The space between the plates is filled with a dielectric material with a dielectric constant \(\kappa_2 = 5\).

The new capacitance \(C_2\) will be:

\(C_2 = \frac{\kappa_2 \epsilon_0 A}{d_2}\)

Calculating the New Capacitance

Now, substitute the values of \(\kappa_2\) and \(d_2\) into the equation for \(C_2\):

\(C_2 = \frac{5 \epsilon_0 A}{d_1/2}\)

We can simplify this expression:

\(C_2 = \frac{5 \cdot 2 \cdot \epsilon_0 A}{d_1}\)

\(C_2 = 10 \cdot \frac{\epsilon_0 A}{d_1}\)

From the initial state, we know that \(\frac{\epsilon_0 A}{d_1} = 6 \text{ pF}\). Substitute this value into the equation for \(C_2\):

\(C_2 = 10 \cdot (6 \text{ pF})\)

\(C_2 = 60 \text{ pF}\)

So, the new capacitance when the distance is halved and the space is filled with a dielectric constant 5 is 60 pF.

Conclusion

Reducing the distance between the plates of a parallel plate capacitor increases the capacitance (inversely proportional to distance). Introducing a dielectric material with a constant κ also increases the capacitance (directly proportional to κ). In this case, the distance reduction by a factor of 2 and the introduction of a dielectric with \(\kappa=5\) collectively increase the capacitance by a factor of \(5 \times 2 = 10\).

Initial Capacitance = 6 pF

Final Capacitance = 10 \(\times\) Initial Capacitance = 10 \(\times\) 6 pF = 60 pF.

The calculated capacitance is 60 pF.

Capacitance Revision Table

Additional Information on Capacitors and Dielectrics

Capacitors are essential components in electronic circuits used for storing electrical energy in an electric field. The amount of charge a capacitor can store at a given voltage is its capacitance.

• Permittivity (\(\epsilon\)): This is a measure of how an electric field affects, and is affected by, a dielectric medium. Higher permittivity means the material is more effective at storing electrical energy for a given electric field.
• Dielectric Strength: While dielectric constant tells us how much capacitance increases, dielectric strength is the maximum electric field a dielectric material can withstand before it breaks down and conducts electricity.
• Energy Stored in a Capacitor: The energy \(U\) stored in a capacitor is given by \(U = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{1}{2} \frac{Q^2}{C}\), where \(Q\) is the charge and \(V\) is the voltage.
• Factors Affecting Capacitance: As seen in the problem, the capacitance of a parallel plate capacitor depends solely on its physical geometry (plate area and separation) and the material between the plates (dielectric constant). It does not depend on the voltage applied or the charge stored.