Problem 168 - A Complete Resource Book in Physics for JEE Main

Question

Let \(P(r)=\frac{Q}{\pi R^{4}} r\) be the charge density distribution for a solid sphere of radius \(R\) and total charge \(Q .\) For a point \(P\) inside the sphere at distance \(r_{i}\) from the centre of the sphere, the magnitude of electric field is (A) Zero (B) \(\frac{Q}{4 \pi \varepsilon_{0} r_{1}^{2}}\) (C) \(\frac{Q r_{1}^{2}}{4 \pi \varepsilon_{0} R^{4}}\) (D) \(\frac{Q r_{1}^{2}}{3 \pi \varepsilon_{0} R^{4}}\)

Step-by-Step Solution

Answer

The short answer is: (C) \(\frac{Q r_{1}^{2}}{4 \pi \varepsilon_{0} R^{4}}\)

Step 1: Calculate total charge within the sphere having radius r_1

First, we need to find the total charge enclosed within the sphere of radius r_1. To do this, we'll integrate the charge density distribution function P(r) over the volume of the inner sphere. The total charge (Q_1) can be calculated using the following formula: \[Q_1 = \int_0^{r_1}\int_0^{\pi}\int_0^{2\pi} \rho(r)r^2\sin(\theta) dr d\theta d\phi\] Since \(P(r) = \frac{Q}{\pi R^4}r\), we have: \[Q_1 = \int_0^{r_1}\int_0^{\pi}\int_0^{2\pi} \frac{Q}{\pi R^4}r \cdot r^2 \sin(\theta) dr d\theta d\phi\]

Step 2: Perform integration

Next, perform the integration w.r.t r, θ, and φ. \[Q_1 = \frac{Q}{\pi R^4} \int_0^{r_1} r^3 dr \cdot \int_0^{\pi} \sin(\theta) d\theta \cdot \int_0^{2\pi} d\phi \] After performing each integration, we get: \[Q_1 = \frac{Q}{\pi R^4} \cdot \left[\frac{r_1^4}{4}\right] \cdot [-\cos(\pi) - 0] \cdot[ 2\pi - 0] \] Simplifying the expression: \[ Q_1 = \frac{Q r_1^4}{2R^4} \]

Step 3: Apply Gauss's Law and find electric field

Now that we have the total charge within the sphere of radius r_1, we can apply Gauss's Law to find the electric field (E) at point P: \[ \oint E \cdot dA = \frac{Q_{1_{inside}}}{\varepsilon_0} \] Since the electric field is radially outward and uniform on the Gaussian surface, we can write: \[E \cdot 4\pi r_1^2 = \frac{Q_1}{\varepsilon_0}\] Now, plug in the expression of Q_1 obtained earlier: \[E \cdot 4\pi r_1^2 = \frac{\frac{Qr_1^4}{2R^4}}{\varepsilon_0}\]

Step 4: Solve for electric field E

Now, we will solve for the electric field E at point P: \[E = \frac{Q r_1^2}{8\pi \varepsilon_0 R^{4}}\cdot 2\] \[E = \frac{Q r_1^2}{4\pi \varepsilon_0 R^{4}}\] Comparing the obtained formula with the given options, we can see that the correct answer is: (C) \(\frac{Q r_{1}^{2}}{4 \pi \varepsilon_{0} R^{4}}\)

Key Concepts

Charge Density Distribution

The concept of charge density distribution is pivotal in understanding how charge is spread out within an object. In the case of our solid sphere, the charge density \( P(r) \) is given as a function of the spherical radius \( r \) and is proportional to the distance from the sphere's center. To grasp the influence of this distribution on the internal electric field, we integrate the charge density over the sphere's volume.

With a non-uniform density, like \( P(r) = \frac{Q}{\pi R^{4}} r \), the charge is not evenly spread throughout the sphere. Instead, more charge is concentrated towards the outer regions compared to near the center. This varying distribution affects the resulting electric field at any point inside the sphere. Understanding this will enable students to approach similar problems involving non-constant charge distributions efficiently.

Gauss's Law

Gauss's Law is a fundamental principle that relates the electric field emanating from a closed surface to the charge enclosed within it. In essence, it provides a method to calculate electric fields for symmetrical charge distributions without complex integration at every point. For a spherical distribution, we consider a Gaussian surface - an imaginary sphere - centered at the same point as the charge distribution.

According to Gauss's Law, the electric flux through the surface is proportional to the charge within the surface, divided by the permittivity of free space (\(\varepsilon_0\)). In symbols, this translates to \(\oint E \cdot dA = \frac{Q_{inside}}{\varepsilon_0}\). Knowledge of how to apply Gauss's Law is crucial here, allowing the student to simplify the problem into a manageable form. The sphericity and symmetry of the charge distribution around the center make Gauss's Law exceptionally suitable for finding the electric field in this scenario.

Integration in Spherical Coordinates

Integration in spherical coordinates becomes important when dealing with problems in three dimensions involving spheres. This coordinate system breaks down a volume into radial distance (\(r\)), polar angle (\(\theta\)), and azimuthal angle (\(\phi\)). For a solid sphere, we integrate the charge density over the volume using the limits of these coordinates. The integration factor \(r^2\sin(\theta)\) compensates for the volume element in spherical coordinates.

Hence, we see an equation like \(Q_1 = \int_0^{r_1}\int_0^{\pi}\int_0^{2\pi} \rho(r)r^2\sin(\theta) dr d\theta d\phi\) representing the total charge within a sphere of radius \(r_1\). Here, integrating over \(\theta\) and \(\phi\) encompasses all directions, while the integration over \(r\) from 0 to \(r_1\) ensures we capture the varying charge density up to the point of interest inside the sphere. Mastering integration in spherical coordinates is vital for students as it is widely used in physics and engineering to solve problems involving three-dimensional symmetry.