Problem 72 - Master Resource Book in JEE Main Physics

Question

The electrostatic potential inside a charged spherical ball is given by \(\phi=a r^{2}+b\), where, \(r\) is the distance from the centre, \(a\) and \(b\) are constants. Then the charge density inside the ball is (a) \(-24 \pi a \varepsilon_{0} r\) (b) \(-6 a \varepsilon_{0}\) (c) \(-24 \pi \varepsilon_{0}\) (d) \(-6 a \varepsilon_{0} r\)

Step-by-Step Solution

Answer

The charge density inside the ball is \(-6a \varepsilon_0\).

Step 1: Recall Gauss's Law for Electrostatics

Gauss's Law relates the electric field to the charge density. For a volume charge density \( \rho \), the law states that the charge density is \( \rho = \varepsilon_0 abla \cdot \mathbf{E} \), where \( \mathbf{E} \) is the electric field and \( \varepsilon_0 \) is the permittivity of free space.

Step 2: Relate Electric Field to Potential

In electrostatics, the electric field \( \mathbf{E} \) is the negative gradient of the potential \( \phi \). Thus, \( \mathbf{E} = -abla \phi \). Given \( \phi = a r^2 + b \), calculate \( abla \phi \) in spherical coordinates: \( abla \phi = \frac{d \phi}{d r} = 2ar \). Hence, \( \mathbf{E} = -2ar \hat{r} \).

Step 3: Calculate Divergence of the Electric Field

To find the divergence of \( \mathbf{E} \), use the formula \( abla \cdot \mathbf{E} = \frac{1}{r^2} \frac{d}{dr} (r^2 E_r) \). Substitute \( E_r = -2ar \), which gives \( abla \cdot \mathbf{E} = \frac{1}{r^2} \frac{d}{dr} (-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a \).

Step 4: Use Gauss's Law to Find Charge Density

Now substitute the divergence \( abla \cdot \mathbf{E} = -6a \) into Gauss's Law: \( \rho = \varepsilon_0 abla \cdot \mathbf{E} = \varepsilon_0 (-6a) = -6a \varepsilon_0 \).

Step 5: Match with Given Options

Compare \( \rho = -6a \varepsilon_0 \) with the options provided. It matches option (b): \(-6a \varepsilon_0 \).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that provides a connection between the electric field and charge distribution. This law states that the electric flux through a closed surface is directly proportional to the total charge enclosed within that surface. It is expressed mathematically as:\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \oint \mathbf{E} \cdot d\mathbf{A} \) represents the electric flux through a closed surface, \( Q_{enc} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.
Gauss's Law is particularly useful for calculating electric fields when dealing with symmetric charge distributions such as spherical, cylindrical, or planar symmetries.
In the context of our problem, Gauss's Law helps relate the divergence of the electric field to the charge density within a spherical region. By using this law, we can accurately determine the charge density given the electric field.

Charge Density

Charge density is a measure of electric charge per unit volume in a region of space. It indicates how much charge is present in a specific area, and is a crucial quantity in understanding electrostatic systems.
In this exercise, we are concerned with volume charge density, represented by \( \rho \), which is measured as charge per cubic meter (C/m³).
Utilizing Gauss's Law, the charge density can be found using the formula:\[ \rho = \varepsilon_0 abla \cdot \mathbf{E} \]This equation states that the charge density is equal to the permittivity of free space \( \varepsilon_0 \) times the divergence of the electric field \( \mathbf{E} \).
The exercise solution showcases this relationship. The divergence of the electric field is computed resulting in a value of \(-6a\), which leads to calculating the charge density as \( -6a \varepsilon_0 \). Matching this result to the provided options gives the correct answer. Understanding charge density allows us to quantify the distribution of charge within the spherical ball in this problem.

Electric Field

The electric field \( \mathbf{E} \) is a vector field that represents the force a charge would experience at any given point in space due to other charges. In electrostatics, it is related to the electric potential \( \phi \) by the negative gradient:\[ \mathbf{E} = -abla \phi \]Given the potential function \( \phi = a r^2 + b \) for a spherical charge distribution, the task involves finding \( abla \phi \).
Calculating this gradient in spherical coordinates yields:\[ abla \phi = \frac{d \phi}{d r} = 2ar \]Thus, the electric field is:\[ \mathbf{E} = -2ar \hat{r} \]where \( \hat{r} \) denotes the radial unit vector.
This negative sign indicates the direction of the field is towards decreasing potential. Knowing how to derive the electric field from a given potential is essential in solving electrostatic problems.
In this scenario, determining \( \mathbf{E} \) was a pivotal step in finding the charge density, showcasing the interconnectedness of electric field, potential, and charge distribution. By comprehending these relationships, electrostatic problems become more intuitive to solve.