Problem 180 - A Complete Resource Book in Physics for JEE Main

Question

A parallel plate capacitor is made of two circular plates separated by a distance of \(5 \mathrm{~mm}\) and with a dielectric constant of \(2.2\) between them. When the electric field in the dielectric is \(3 \times 10^{4} \mathrm{~V} / \mathrm{m}\), the charge density of the positive plate will be close to (A) \(6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}\) (B) \(3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}\) (C) \(3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}\) (D) \(6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

The charge density of the positive plate is closest to (A) \(6 \times 10^{-7}~C/m^2\).

Step 1: Write down the given values

We are given: - Distance between plates (d) = 5 mm = \(5 \times 10^{-3} m\) - Dielectric constant (κ) = 2.2 - Electric field (E) = \(3 \times 10^{4}~V/m\)

Step 2: Determine the voltage

Use the formula for the electric field to find the voltage: \[E = \frac{V}{d}\] Rearrange the formula to solve for V: \[V = E \times d\] Now substitute the given values: \[V = (3 \times 10^{4}) \times (5 \times 10^{-3})\] \[V = 150~V\]

Step 3: Determine the capacitance

Use the formula for a parallel plate capacitor to determine the capacitance (C): \[C = \frac{κε_0A}{d}\] We don't have the surface area (A) of the plate in this equation. However, as we can see, in the formula for charge density, the area would get canceled out, hence we don't need to find it. We know that \(ε_0 = 8.85 \times 10^{-12}~F/m\)

Step 4: Determine the charge density

Now use the relationship between the charge density (σ), capacitance (C), and the dielectric constant (κ): \[σ = \frac{CE}{κ}\] Substitute the values: \[σ = \frac{(\frac{κε_0A}{d})E}{κ}\] Area (A) and the dielectric constant (κ) cancels out, leaving: \[σ = \frac{ε_0AE}{d}\] Remembering we don't need the exact value of A, we can determine the charge density only by substituting the given values: \[σ = \frac{(8.85 \times 10^{-12})(3 \times 10^{4})}{(5 \times 10^{-3})}\] \[σ = 5.31 \times 10^{-7}~C/m^2\] Comparing this result to the given options, we can see that it is closest to the value in option (A): \[6 \times 10^{-7}~C/m^2\] Therefore, the correct answer is (A) \(6 \times 10^{-7}~C/m^2\).

Key Concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, representing the force per unit charge that a charged object would experience in the vicinity of other charges or due to a potential difference between two points. For a parallel plate capacitor, which consists of two charged plates separated by an insulating material or vacuum, the electric field (\(E\)) is uniform and directed from the positive plate to the negative one. It is calculated using the formula \[ E = \frac{V}{d} \] where \(V\) is the voltage across the plates and \(d\) is the distance between them. Understanding the electric field is crucial because it directly affects the charge distribution across the capacitor's plates and thus influences the charge density, which is a measure of how much electric charge is accumulated per unit area.

Dielectric Constant

The dielectric constant, also known as the relative permittivity (\(\kappa\)), is a measure of a material's ability to pass an electric field through it. Specifically, it quantifies how much the material reduces the electric field compared to a vacuum, which has a dielectric constant of 1. In the context of a parallel plate capacitor, inserting a dielectric material between the plates increases the capacitance because the dielectric reduces the effective electric field inside the capacitor. This relationship helps one to understand why, in the given step-by-step solution, the capacitance formula \[ C = \frac{\kappa\epsilon_0A}{d} \] incorporates the dielectric constant—higher \(\kappa\) allows for greater charge storage \(Q\) for the same voltage \(V\), following the equation \(Q = CV\), and thus affects the charge density on each plate. Additionally, the dielectric constant also dictates how much voltage a capacitor can withstand before it fails, or the dielectric breaks down.

Charge Density

Charge density (\(\sigma\)) is the measure of electric charge per unit area on a surface. In the case of a parallel plate capacitor, the charge density on the surface of one plate is directly related to the amount of electric charge \(Q\) it holds, and inversely proportional to its surface area \(A\): \[ \sigma = \frac{Q}{A} \] When dealing with a parallel plate capacitor that has a dielectric inserted, the charge density on the plates is given by the formula: \[ \sigma = \frac{\epsilon_0AE}{d} \] where \(\epsilon_0\) is the vacuum permittivity, and once again, \(A\) gets canceled out when calculating the charge density, particularly in the context of this problem where the actual area isn't given. Charge density is significant for understanding the behavior and limitations of capacitors, including how much energy they can store and how they might behave under different charging scenarios. The charge density along with the dielectric constant can indicate the maximum charge a capacitor can hold before dielectric breakdown occurs. Ultimately, it affects the device's stability and performance.