Problem 125 - A Complete Resource Book in Physics for JEE Main

Question

The energy density in the electric field created by a point charge falls off with distance from the point charge as (A) \(\frac{1}{r}\) (B) \(\frac{1}{r^{2}}\) (C) \(\frac{1}{r^{3}}\) (D) \(\frac{1}{r^{4}}\)

Step-by-Step Solution

Answer

The energy density in the electric field created by a point charge falls off with distance from the point charge as \(\frac{1}{r^4}\) (option D). This can be determined by substituting the formula for the electric field due to a point charge, \(\frac{kQ}{r^2}\), into the formula for energy density, \(\frac{1}{2} \epsilon_0 E^2\), and simplifying the expression.

Step 1: Recall the formula for electric field due to a point charge

The electric field, \(E\), due to a point charge, \(Q\), at a distance \(r\) from the charge is given by Coulomb's Law: \[E = \frac{kQ}{r^2}\] where \(k\) is the electrostatic constant, approximately \(8.99 \times 10^{9} \frac{N m^2}{C^2}\).

Step 2: Recall the formula for energy density in an electric field

The energy density, \(u\), in an electric field, is given by the following formula: \[u = \frac{1}{2} \epsilon_0 E^2\] where \(\epsilon_0\) is the vacuum permittivity, approximately \(8.85 \times 10^{-12} \frac{C^2}{N m^{2}}\), and \(E\) is the electric field.

Step 3: Substitute the formula for the electric field in the energy density formula

We substitute the expression for the electric field due to a point charge into the formula for energy density: \[u = \frac{1}{2} \epsilon_0 \left(\frac{kQ}{r^2}\right)^2\]

Step 4: Simplify the expression for energy density

Now, simplify the expression by squaring the electric field term and combining constants: \[u = \frac{1}{2} \epsilon_0 \frac{k^2 Q^2}{r^4}\]

Step 5: Determine how the energy density falls off with distance

We can see from the simplified expression that the energy density, \(u\), falls off with distance, \(r\), as \(\frac{1}{r^4}\). Therefore, the correct answer is (D) \(\frac{1}{r^{4}}\).