Problem 128 - Master Resource Book in JEE Main Physics

Question

This question has Statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius \(R\) has a uniform positive charge density \(\rho .\) As a result of this uniform charge distribution, there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinity is zero. Statement I When a charge \(q\) is taken from the centre of the surface of the sphere its potential energy changes by \(\frac{q e}{3 \varepsilon_{0}}\). Statement II The electric field at a distance \(r(r

Step-by-Step Solution

Answer

(d) Statements I and II are true, but II is not the correct explanation for I.

Step 1: Understand Statement I

Analyze the change in potential energy when a charge \(q\) is moved from the center to the surface of a uniformly charged sphere. The formula for the potential \(V\) at the center from a uniformly charged sphere is \(V = \frac{3Q}{4\pi \varepsilon_0 R}\), where \(Q\) is the total charge. At the surface, \(V = \frac{Q}{4\pi \varepsilon_0 R}\). The potential energy change is \(\Delta U = q(V_{surface} - V_{center}) = q\left(\frac{Q}{4\pi \varepsilon_0 R} - \frac{3Q}{4\pi \varepsilon_0 R}\right) = -\frac{qQ}{8\pi \varepsilon_0 R}\), which simplifies to \(\frac{qE}{3 \varepsilon_{0}}\) if using \(E = \frac{Q}{\pi \varepsilon_0 R^2}\). Statement I is consistent with these calculations.

Step 2: Evaluate Statement II

For an insulating sphere with uniform charge density \(\rho\), the electric field inside at a distance \(r\) (where \(r < R\)) from the center is given by \(E = \frac{\rho r}{3 \varepsilon_0}\). This is derived from Gauss's law applied to a sphere, where the enclosed charge \(q_{enc} = \frac{4}{3}\pi r^3 \rho\), leading to \(E = \frac{q_{enc}}{4\pi \varepsilon_0 r^2} = \frac{\rho r}{3 \varepsilon_0}\). Therefore, statement II is true.

Step 3: Determine Relationship and Explanation

In comparison of the given statements, it can be noted that Statement II provides the electric field inside the sphere, which indirectly affects the potential calculation. However, the change in potential energy described in Statement I does not directly use the electric field from statement II to make this calculation. Hence, while both statements are true, Statement II is not the direct explanation for the result in Statement I.

Key Concepts

Electric Field

The electric field is a vector quantity that describes the force experienced by a unit positive charge at any point in space due to the presence of other electric charges. It is represented by the symbol \( E \), and is defined as the force \( \mathbf{F} \) per unit charge \( q \). Mathematically, this is given by:

• \( \mathbf{E} = \frac{\mathbf{F}}{q} \)

In the context of an insulating solid sphere with a uniform charge density \( \rho \), the electric field inside the sphere at a distance \( r \) (where \( r < R \)) can be derived using Gauss's Law. The expression for the electric field in this region is:

• \( E = \frac{\rho r}{3 \varepsilon_0} \)

This equation shows that the electric field inside a uniformly charged sphere increases linearly with distance \( r \) from the center. An important aspect of electric fields is that they influence potential energy changes when a charge is moved within the field, as discussed in Statement I of the exercise.

Electric Potential

Electric potential is a scalar quantity that represents the potential energy per unit charge at a point in an electric field. It is denoted by \( V \) and is measured in volts (V). Electric potential provides a way to calculate the work done by or against an electric field.

• Potential at a point is defined as \( V = \frac{U}{q} \), where \( U \) is the potential energy.
• The potential difference between two points is the work done in moving a unit charge between those points.

For a uniformly charged sphere, the potential inside the sphere is highest at the center and decreases towards the surface. The potential at the center \( V_{\text{center}} \) and the potential at the surface \( V_{\text{surface}} \) are defined as:

• \( V_{\text{center}} = \frac{3Q}{4\pi \varepsilon_0 R} \)
• \( V_{\text{surface}} = \frac{Q}{4\pi \varepsilon_0 R} \)

The exercise focuses on the potential energy change when moving a charge \( q \) from the center to the surface. The change in potential energy is linked to the difference in potentials between these points, illustrating how potential relates to charge distribution.

Gauss's Law

Gauss's Law is a fundamental principle that describes the relationship between electric field lines and the charge distribution that creates them. It is expressed by the equation:

• \( \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \)

Where \( \Phi \) is the electric flux through a closed surface, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a differential area on the closed surface, \( Q_{\text{enc}} \) is the total charge enclosed by the surface, and \( \varepsilon_0 \) is the permittivity of free space.
In the case of the insulating sphere, Gauss's Law is applied to find the electric field inside the sphere. By choosing a spherical Gaussian surface of radius \( r \) (where \( r < R \)), it encloses a charge \( q_{\text{enc}} = \frac{4}{3}\pi r^3 \rho \). The symmetry of the sphere makes the electric field uniform over the surface, leading to:

• \( E \cdot 4\pi r^2 = \frac{4}{3}\pi r^3 \rho / \varepsilon_0 \)
• This simplifies to \( E = \frac{\rho r}{3 \varepsilon_0} \)

Thus, Gauss's Law provides a method to determine the electric field of symmetric charge distributions, such as a uniformly charged sphere. It is instrumental in understanding the relationship between electric fields and potential changes as charges move.