Question

A parallel plate capacitor with square plates of edge length \(L\) separated by a distance \(d\) is given a charge \(Q\), then disconnected from its power source. A close-fitting square slab of dielectric, with dielectric constant \(\kappa\), is then inserted into the previously empty space between the plates. Calculate the force with which the slab is pulled into the capacitor during the insertion process.

Short answer

Answer: The force acting on the dielectric slab during the insertion process is given by \(F = \frac{\Delta U}{d}\), where \(\Delta U\) is the change in stored energy due to the insertion of the slab, and \(d\) is the distance between the capacitor plates.

Step-by-step solution

Calculate the initial capacitance

Before the dielectric slab is inserted into the capacitor, the capacitance can be calculated using the formula: \(C_0 = \frac{\epsilon_0 A}{d}\) where \(A = L^2\) is the area of the plates, and \(\epsilon_0\) is the vacuum permittivity.

Calculate the initial potential difference

The potential difference across the capacitor before the slab is inserted is given by the formula: \(V_0 = \frac{Q}{C_0}\)

Calculate the initial stored energy

The energy stored in the capacitor before the slab is inserted can be calculated using the formula: \(U_0 = \frac{1}{2} C_0 V_0^2\)

Calculate the final capacitance

After the dielectric slab is inserted into the capacitor, the formula for the capacitor is now: \(C_1 = \kappa C_0\)

Calculate the final potential difference

Since the capacitor is disconnected from the power source and the charge remains constant, the potential difference decreases after the insertion of the dielectric slab: \(V_1 = \frac{Q}{C_1} = \frac{Q}{\kappa C_0}\)

Calculate the final stored energy

The energy stored in the capacitor after the slab is inserted can be calculated with the same formula as before: \(U_1 = \frac{1}{2} C_1 V_1^2\)

Calculate the difference in stored energy

The change in stored energy due to the insertion of the dielectric slab is given by: \(\Delta U = U_1 - U_0\)

Determine the force acting on the dielectric slab

The force acting on the dielectric slab during the insertion process can be found using the fact that the work done is equal to the change in stored energy. The work done is given by the product of force and distance; therefore, \(F = \frac{\Delta U}{d}\)

Key concepts

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit of potential difference across its plates. It is a fundamental concept in understanding how capacitors function and is denoted by the symbol C. For a parallel plate capacitor, without any dielectric material, the capacitance is given by the equation:
\[C = \frac{\epsilon_0 A}{d}\]
where \(\epsilon_0\) is the permittivity of free space, \(A\) is the area of one of the plates, and \(d\) is the distance between the two plates. As we see in the problem, when the dimensions of the plates and the separation between them are known, one can easily calculate the initial capacitance, which is crucial for subsequent steps.

Dielectric Constant

The dielectric constant, often symbolized by \(\kappa\), is a dimensionless measure of a material's ability to increase the capacitance of a capacitor when placed between the plates. It is also known as the relative permittivity of the dielectric material.

Role of Dielectric in Capacitors

When a dielectric is inserted into a capacitor, as in our exercise, the capacitance increases according to the formula:
\[C_1 = \kappa C_0\]
This relationship reveals that the dielectric constant is a multiplier that enhances the original capacitance, \(C_0\), to a new value, \(C_1\). The higher the dielectric constant, the greater the capacitance and thus the greater the charge the capacitor can store at the same potential difference.

Potential Difference

The potential difference, or voltage, across the plates of a capacitor is the amount of electric potential energy per unit charge. It reflects the energy required to move a charge between the two plates and is a key factor in determining the stored energy within the capacitor.
In our example, the initial voltage is given by:
\[V_0 = \frac{Q}{C_0}\]
After inserting the dielectric slab, the new potential difference decreases due to the increased capacitance:
\[V_1 = \frac{Q}{C_1}\]
This reduction in voltage highlights the impact of the dielectric on the overall electric field and potential difference within the capacitor.

Stored Energy in Capacitors

The stored energy in a capacitor is the amount of electrical energy held within it, which can be calculated when the capacitance and the voltage are known. The energy is found using the formula:
\[U = \frac{1}{2} C V^2\]
This equation demonstrates that the energy stored is proportional to the square of the voltage and directly proportional to the capacitance. As shown in the steps of the exercise, the initial energy is calculated before the slab insertion, and the final energy is computed after the slab is in place. The difference in energy is significant because it represents the work done by the electric field on the dielectric material during its insertion.

Force on Dielectric

The force exerted on a dielectric when it is inserted into a capacitor can be understood by considering the change in energy within the system. This force, which pulls the dielectric into the capacitor, can be found by dividing the change in stored energy by the distance the dielectric moves:
\[F = \frac{\Delta U}{d}\]
In the context of our exercise, the dielectric experiences a force due to the work done by the electric field to pull it into the capacitor. The work done is equivalent to the change in stored energy, which is calculated by subtracting the initial energy from the final energy after the dielectric is fully inserted. By applying this force equation, students can quantitatively analyze the effect of inserting a dielectric into a capacitor, a pivotal concept in the study of capacitors and their behavior in electric circuits.