Problem 89 - Master Resource Book in JEE Main Physics

Question

A parallel plate capacitor of plate area \(A\) and plate separation \(d\) is charged to potential difference \(V\) and then the battery is disconnected. A slab of dielectric constant \(K\) is then inserted between the plates of the capacitor so as to fill the space between the plates. If \(Q, E\) and \(W\) denotes respectively the magnitude of charges on each plate, the electric field between the plates (after the slab is inserted) and work done on the system, in the process of inserting the slab, then (a) \(Q=\frac{\varepsilon_{0} A V}{d}\) (b) \(Q=\frac{\varepsilon_{0} K A V}{d}\) (c) \(E=\frac{V}{K d}\) (d) \(W=\frac{\varepsilon_{0} A V^{2}}{2 d}\left[1-\frac{1}{K}\right]\)

Step-by-Step Solution

Answer

(a) Q=\(\frac{\varepsilon_0 A V}{d}\), (c) E=\(\frac{V}{K d}\), (d) W=\(\frac{\varepsilon_0 A V^2}{2d}\times\left[1-\frac{1}{K}\right]\).

Step 1: Charge on Plates (Q) after Inserting Dielectric

The initial charge on the plates when the capacitor is charged to a potential difference \(V\) is given by \(Q = C \cdot V = \frac{\varepsilon_0 A V}{d}\). When a dielectric is inserted, the charge remains constant since the battery is disconnected. Therefore, the correct expression for the charge \(Q\) after the dielectric is inserted is \(Q = \frac{\varepsilon_0 A V}{d}\).

Step 2: Electric Field (E) between Plates after Insertion

The electric field \(E\) in a capacitor after inserting a dielectric is given by \(E = \frac{E_0}{K}\), where \(E_0\) was the initial electric field \(E_0 = \frac{V}{d}\). Thus, after insertion, the electric field is \(E = \frac{V}{K d}\).

Step 3: Work Done (W) on the System

The work done on the system when a dielectric slab is inserted can be calculated using the energy change. Initial energy is \(U_i = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\). After inserting the dielectric, the energy becomes \(U_f = \frac{1}{2} \frac{\varepsilon_0 A V^2}{K d}\). The work done \(W\) by the system is the difference \(W = U_i - U_f = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)\).

Key Concepts

Dielectric Materials

Dielectric materials are insulating substances that do not conduct electricity but can support an electric field. When inserted into a capacitor, they have amazing effects. They increase the capacitor's ability to store charge by reducing the effective electric field within. This phenomenon is because these materials polarize in response to the electric field, reducing the field's strength between the plates.
To understand how a dielectric works, consider how it changes the permittivity within the capacitor. The dielectric constant, denoted by the symbol \(K\), measures a material's ability to improve a capacitor's capacitance. A higher dielectric constant means more charge can be stored at the same voltage.

• Improves capacitance: Allows the capacitor to store more energy for the same amount of voltage
• Reduces electric field strength: Decreases the electric field between the plates, making the capacitor more efficient
• Stabilizes voltage fluctuations: By allowing stored energy to be held, it provides a smoother voltage output

When applying these ideas, the original charge, \(Q\), on the capacitor does not change after inserting a dielectric since the battery is disconnected. The dielectric merely modifies the other properties, leading to new capacitance and field behaviors.

Electric Field

An electric field in a capacitor is the region between the plates where electrical forces can act. This field is pivotal in determining how the capacitor will function with or without a dielectric present. Without a dielectric, the electric field \(E_0\) can be described by the equation \(E_0 = \frac{V}{d}\), where \(d\) is the separation between the plates.
When a dielectric is inserted, the electric field is altered. The dielectric reduces the effective field strength by a factor of the dielectric constant \(K\). This can be demonstrated by the formula \(E = \frac{V}{K d}\). This reduced field has several practical implications:

• Reduces energy leakage: A lower electric field reduces the potential for external influences to sap energy from the capacitor
• Enhances energy storage: By reducing internal field strength, capacitors can handle more energy without breakdown

Understanding how dielectrics affect electric fields helps in the design of more efficient capacitors and electronic circuits, optimizing the use of space and materials for energy storage and management.

Energy in Capacitors

Capacitors are essential components for storing energy in electrical systems. The energy stored in a capacitor without a dielectric is given by \(U_i = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d}\). This equation states that the energy is proportional to the square of the voltage and inversely proportional to the distance between the plates.
When a dielectric is introduced, the stored energy changes, leading to a different final energy \(U_f = \frac{1}{2} \frac{\varepsilon_0 A V^2}{K d}\). The reduction of energy due to the dielectric is equivalent to the work done on the system. The work \(W\) can be described as \(W = U_i - U_f = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)\). This equation helps us realize that:

• Work is performed in altering energy: Energy adjustments with dielectrics involve work even if the battery is not connected
• Energy efficiency: Dielectric material improves the storage efficiency
• System stabilization: By managing energy levels efficiently, it helps stabilize power supply systems

Recognizing how capacitors manage energy with dielectrics is vital for effective design and functioning of electronic devices.