Problem 86 - A Complete Resource Book in Physics for JEE Main

Question

A parallel plate capacitor of capacitance \(C\) is connected to a battery of \(e m f V\). If a dielectric slab is completely inserted between the plates of the capacitor and battery remains connected, then electric field between plates (A) decreases. (B) increases. (C) remains constant. (D) may be increase or may be decrease.

Step-by-Step Solution

Answer

The electric field between the plates remains constant when a dielectric slab is completely inserted between the plates of the capacitor and the battery remains connected. The correct answer is (C) Remains constant.

Step 1: Find the initial electric field without the dielectric slab

Initially, the capacitor is connected to a battery with an emf V, and there is no dielectric slab inserted. We will use the formula for capacitance, which is given by \(C = \frac{Q}{V}\), where Q is the charge stored on the capacitor plates and V is the potential difference across the plates. Now, we know that the electric field, E, between the parallel plates of the capacitor is given by the formula \(E = \frac{V}{d}\), where d is the distance between the plates. To find the electric field without the dielectric slab, we will use this formula.

Step 2: Find the charge stored on the capacitor plates

Using the formula for capacitance, \(C = \frac{Q}{V}\), we can find the charge stored on the plates of the capacitor as follows: \[Q = CV\]

Step 3: Inserting the dielectric slab

When the dielectric slab is inserted between the plates of the capacitor, the capacitance increases by a factor of the dielectric constant, K. So, the new capacitance becomes \(C' = KC\). Since the battery is still connected, the potential difference across the capacitor remains the same; hence, V is constant. Now, we can find the new charge on the plates using the formula: \[Q' = C'V\] Substituting the new capacitance value, we get: \[Q' = KCV\]

Step 4: Find the new electric field with the dielectric slab

The new electric field between the plates with the dielectric slab can be found using the formula \(E' = \frac{Q'}{Kd}\), where E' is the new electric field and Q' is the new charge on the plates. Substituting the expression for \(Q'\) from step 3, we get: \[E' = \frac{KCV}{Kd}\] Now, since \(E = \frac{V}{d}\), we can rewrite the expression for E' as: \[E' = \frac{KE}{K}\]

Step 5: Compare the initial and new electric fields

Now we can see the relationship between the electric fields before and after the insertion of the dielectric slab: \[E' = \frac{KE}{K}\] Simplifying this expression, we find that: \[E' = E\] This means that the electric field between the plates remains constant after the insertion of the dielectric slab while the battery is connected. So, the correct answer is: (C) Remains constant.

Key Concepts

Capacitance

Capacitance is a property of a capacitor which measures its ability to store electrical energy in an electric field. The capacitance, denoted by the symbol \(C\), is defined as the ratio of the electric charge \(Q\) on one conductor to the potential difference \(V\) between the two conductors, and is given by the equation \(C = \frac{Q}{V}\).

In the context of a parallel plate capacitor, the capacitance is directly proportional to the area of the plates \(A\) and inversely proportional to the distance \(d\) between them. The medium between the plates also affects capacitance; the introduction of a dielectric material, for instance, increases the overall capacitance of the system. The formula for capacitance in a vacuum or air is \(C = \frac{\text{ε}_0A}{d}\), where \text{ε}_0 is the permittivity of free space. In short, capacitance indicates how much charge a capacitor can store at a given voltage.

Dielectric Slab

A dielectric slab is an insulating material that can be placed between the plates of a capacitor to increase its capacitance. Dielectrics are characterized by their dielectric constant \(K\), also known as the relative permittivity. This constant is a measure of how effectively the dielectric material can store electrical energy.

When a dielectric slab is inserted into a capacitor, it reduces the electric field within the capacitor for the same charge on the plates, thereby allowing more charge to be stored at the same voltage. Mathematically, the capacitance of the capacitor with the dielectric slab becomes \(C' = KC\), where \(C\) is the initial capacitance without the dielectric.

This increase in capacitance due to the dielectric material is essential in designing capacitors with higher energy storage capacity without increasing the size of the plates or decreasing the distance between them. A dielectric also has the property of reducing the likelihood of a capacitor's breakdown by increasing the breakdown voltage, enhancing the capacity and safety of the device.

Electric Field

The electric field is a vector quantity that represents the force experienced by a unit positive charge at any point in space. It is a fundamental concept in electromagnetism and is denoted by the symbol \(E\) . In the case of a parallel plate capacitor, the electric field between the plates is uniform and can be calculated using the equation \(E = \frac{V}{d}\), where \(V\) is the potential difference across the plates and \(d\) is the separation between them.

As the exercise suggests, when a dielectric slab is inserted between the plates while the capacitor is connected to a battery, the electric field does not change. This happens because the battery maintains a constant voltage across the plates, and since \(E\) is directly related to \(V\) and inversely to \(d\) , and neither \(V\) nor \(d\) change, \(E\) remains constant. The ability to control the electric field in a capacitor is of great practical importance in electronics, affecting the operation of a wide range of devices including sensors, memories, and filters.