Problem 127 - A Complete Resource Book in Physics for JEE Main

Question

Let \(u_{a}\) and \(u_{d}\) represent the energy density in air and in a dielectric, respectively, for the same field in both. Let \(K=\) dielectric constant. Then (A) \(u_{a}=u_{d}\) (B) \(u_{a}=K u_{d}\) (C) \(u_{d}=K u_{a}\) (D) \(u_{a}=(K-1) u_{d}\)

Step-by-Step Solution

Answer

The correct relationship between the energy density in air (\(u_a\)) and in a dielectric (\(u_d\)) with a dielectric constant (\(K\)) is given by (C) \(u_{d}=K u_{a}\).

Step 1: Determine the formula for energy density in a dielectric

The energy density in a dielectric with a permittivity of \(\epsilon\) and an electric field of \(E\) is given by: \[u = \frac{1}{2} \epsilon E^2\]

Step 2: Determine the relationship between permittivity in air and in a dielectric

The relative permittivity (or dielectric constant) of a dielectric is defined as the ratio of its permittivity to the permittivity of free space (air), which is given by: \[K = \frac{\epsilon_d}{\epsilon_0}\] Where \(\epsilon_d\) is the permittivity in the dielectric, and \(\epsilon_0\) is the permittivity of free space (air).

Step 3: Calculate energy density in the dielectric

By using the relationship between the dielectric constant and permittivity, we can rewrite the energy density in terms of the dielectric constant (\(K\)) and the energy density in the air (\(u_a\)): \[u_d = \frac{1}{2} \epsilon_d E^2 = \frac{1}{2}(\epsilon_0 K) E^2\] Since \(\frac{1}{2} \epsilon_0 E^2 \) is the energy density in the air (\(u_a\)), we can rewrite the equation as follows: \[u_d = K u_a\]

Step 4: Select the correct answer

Now that we have found the relationship between the energy densities in the dielectric and air, we can select the correct answer: (C) \(u_{d}=K u_{a}\)