Problem 135 - 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 \(a, b\) are constants. Then the charge density inside the ball is [AIEEE 2011] (a) \(-6 a \varepsilon_{0} r\) (b) \(-24 \pi a \varepsilon_{0}\) (c) \(-6 a \varepsilon_{0}\) (d) \(-24 \pi a \varepsilon_{0} r\)

Step-by-Step Solution

Answer

The charge density inside the ball is \( \rho = -6a \varepsilon_0 \), which is option (c).

Step 1: Understand the given electrostatic potential

We are given the electrostatic potential inside a charged spherical ball as \( \phi = a r^{2} + b \), where \( r \) is the distance from the center, and \( a \), \( b \) are constants.

Step 2: Apply Poisson's equation in electrostatics

Poisson's equation relates the charge density \( \rho \) to the electric potential \( \phi \). It is given by the formula: \[ abla^{2} \phi = -\frac{\rho}{\varepsilon_{0}} \]. This means we need to calculate the Laplacian \( abla^{2} \phi \) of the potential \( \phi \).

Step 3: Calculate the Laplacian for spherical coordinates

For a potential \( \phi(r) = a r^{2} + b \) in spherical coordinates, the Laplacian \( abla^2 \phi \) is \( abla^{2} \phi = \frac{1}{r^{2}} \frac{d}{dr}\left( r^{2} \frac{d\phi}{dr} \right) \).

Step 4: Compute the first and second derivative of potential

First, compute \( \frac{d\phi}{dr} = \frac{d}{dr}(a r^{2} + b) = 2ar \). Then, we need to compute \( \frac{d}{dr}\left( r^{2} \frac{d\phi}{dr} \right) = \frac{d}{dr}\left( r^{2} \cdot 2ar \right) = \frac{d}{dr}(2ar^3) = 6ar^2 \).

Step 5: Calculate the Laplacian value

Substitute back into the Laplacian formula: \( abla^{2} \phi = \frac{1}{r^{2}} \cdot 6ar^2 = 6a \).

Step 6: Solve for charge density using Poisson's equation

From \( abla^2 \phi = -\frac{\rho}{\varepsilon_0} \), solve for \( \rho \): \( \rho = -6a \varepsilon_0 \). This matches the option (c).

Key Concepts

Poisson's Equation

Poisson's equation is a fundamental concept in electrostatics. It links the electric potential to the charge distribution in a region. This equation is given by:

• \( abla^{2} \phi = -\frac{\rho}{\varepsilon_{0}} \)

where \( \phi \) is the electric potential, \( \rho \) is the charge density, and \( \varepsilon_{0} \) is the permittivity of free space.
In essence, Poisson's equation tells us that the way electric potential changes in a space is directly related to how charge is distributed there. The Laplacian \( abla^{2} \phi \) signifies the way the curvature of the potential \( \phi \) changes, which indicates where more charge is.
This concept is crucial because, in many practical situations, knowing only the electric potential or the charge distribution is not enough. Poisson's equation provides the mathematical link allowing us to determine one from the other. It serves as a powerful tool for solving electrostatic problems, such as calculating electric fields in complex structures.

Charge Density

Charge density \( \rho \) quantifies the amount of electric charge in a given space. In general, it describes how charge is distributed over a particular region within a medium. In this context, we use Poisson's equation to solve for \( \rho \) based on the provided electric potential \( \phi \).
The relationship between charge density and electric potential is such that areas with high charge density will affect the potential significantly, altering the Laplacian or the curvature of \( \phi \).
This exercise involved finding the charge density inside a charged spherical ball by calculating the Laplacian of the potential function \( \phi = a r^{2} + b \) and incorporating it into the Poisson's equation:

• \( \rho = -\varepsilon_{0} abla^{2} \phi \)

Understanding charge density is essential not only in theoretical contexts but also in practical applications, like designing capacitors and understanding how charges accumulate in insulating and conductive materials.

Electric Potential

Electric potential \( \phi \) is a measure of the potential energy per unit charge at a point in an electric field. It provides insight into the distribution of electric potential energy in space, which is influenced by the presence of charge.
In electrostatics, the electric potential can be used to determine many aspects of an electric field and associated forces. For the given spherical charge distribution, the potential function is specific: \( \phi = a r^{2} + b \), where \( a \) and \( b \) are constants, and \( r \) is the radial distance from the center.
Using this potential function, you can derive additional properties of the system, such as the electric field and the force experienced by a charge within the field. Calculating the gradient or the Laplacian of the electric potential is a common task to interpret and solve electrostatic problems.
The relationship between electric potential, electric field, and charge density is interwoven. Electric potential is foundational in determining the physical behavior of systems governed by electrostatic forces, providing a simpler means to calculate work done and energy stored in fields.