This question has Statement I and Statement II. Of the four choices given after the statements, choo

Q: Why did we use dimensional analysis instead of calculating the exact potential difference?

Calculating the exact potential difference requires integrating the electric field from the center to the surface. While that works perfectly, dimensional analysis is a powerful shortcut in competitive exams like JEE. If the dimensions of an expression don't match the physical quantity it represents, the expression is guaranteed to be wrong!

Q: What is the actual change in potential energy?

The potential at the center is $V_c = \frac{3\rho R^2}{6\varepsilon_0}$ and at the surface is $V_s = \frac{2\rho R^2}{6\varepsilon_0}$. The change in potential energy is $\Delta U = q(V_s - V_c) = -\frac{q\rho R^2}{6\varepsilon_0}$.

Q: Why is the electric field inside a solid sphere proportional to $r$?

Inside a uniformly charged solid sphere, the charge enclosed by a Gaussian surface of radius $r$ grows as $r^3$. However, the surface area grows as $r^2$. According to Gauss's Law, $E \propto \frac{q_{\text{enc}}}{r^2} \propto \frac{r^3}{r^2} = r$. Thus, the electric field increases linearly with distance from the center.

Q: Does Gauss's Law apply if the charge density is not uniform?

Yes, Gauss's Law is always valid! However, if the charge density is not uniform (e.g., it depends on $\theta$ or $\phi$), the electric field won't be constant over the Gaussian surface, making it difficult to pull $E$ out of the integral. In such cases, Gauss's Law is less useful for calculating the electric field.

Q: Why is the potential at infinity taken as zero?

Taking the potential at infinity as zero is a standard convention in electrostatics. It provides a convenient reference point because the electric field of a localized charge distribution approaches zero as you move infinitely far away.

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 infinite is zero.
Statement I When a charge $q$ is taken from the centre of the surface of the sphere, its potential energy changes by $\frac{q\rho}{3\varepsilon_0}$.
Statement II The electric field at a distance $r (r < R)$ from the centre of the sphere is $\frac{\rho r}{3\varepsilon_0}$.

Options:

(A) Statement I is false, Statement II is true

(B) Statement I is true, Statement II is false

(C) Statement I is true, Statement II is true; Statement II the correct explanation for Statement I

(D) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a solid insulating sphere of radius $R$. It's packed uniformly with positive charge, meaning its volume charge density, $ho$, is constant throughout. We are given two statements to evaluate. Statement I talks about the change in potential energy when a charge is moved from the center to the surface, and Statement II gives an expression for the electric field inside the sphere.

Let's tackle Statement II first because it's a direct application of a fundamental law.

### The Master Equation: Gauss's Law

To find the electric field at a distance $r < R$, Gauss's Law is our best friend. We draw an imaginary spherical Gaussian surface of radius $r$ inside the sphere.

The charge enclosed by this surface is simply the volume charge density times the volume of this smaller sphere:
$$q_{	ext{enc}} = ho 	imes \left( \frac{4}{3}\pi r^3 ight)$$

Now, we substitute the enclosed charge into Gauss's Law. The electric field is uniform and parallel to the area vector over our Gaussian surface, so the flux is just $E$ times the surface area, $4\pi r^2$:
$$E(4\pi r^2) = \frac{ho \left( \frac{4}{3}\pi r^3 ight)}{\varepsilon_0}$$

Canceling out the common terms, we get the electric field:
$$E = \frac{ho r}{3\varepsilon_0}$$

This exactly matches Statement II! So, Statement II is absolutely **true**.

### The Smart Move: Dimensional Analysis

Now let's look at Statement I. It claims the change in potential energy is $\Delta U = \frac{qho}{3\varepsilon_0}$. Before doing any heavy integration to find the exact potential difference, let's be smart and check its dimensions. Silly mistakes happen here, so pay attention!

Let's break down the units:
- Charge $q$ is in Coulombs ($	ext{C}$)
- Charge density $ho$ is in $	ext{C/m}^3$
- Permittivity $\varepsilon_0$ is in $	ext{C}^2 / (	ext{N}\cdot	ext{m}^2)$

When we plug these units into the expression $\frac{qho}{\varepsilon_0}$, we get:
$$\left[ \frac{qho}{\varepsilon_0} ight] = \frac{	ext{C} \cdot (	ext{C/m}^3)}{	ext{C}^2 / (	ext{N}\cdot	ext{m}^2)} = \frac{	ext{N}}{	ext{m}}$$

The Coulombs cancel out, and we are left with Newtons per meter. But wait! Energy must be in Joules, which is Newton-meters ($	ext{N}\cdot	ext{m}$). The dimensions don't match at all!

### Final Conclusion

Because the dimensions are wrong, Statement I is fundamentally incorrect. We didn't even need to calculate the actual potential difference! So, Statement I is **false**, and Statement II is **true**.

This is a classic example of how dimensional analysis can save you precious time in competitive exams. Always keep an eye on the units!

Select Answer:

Visualized Solution

Visualizing the Charged Sphere

Evaluating Statement II: Gauss's Law

Gaussian Surface and Enclosed Charge

Applying Gauss's Law

Electric Field Inside the Sphere

Evaluating Statement I: Dimensional Analysis

Dimensional Check

Calculating the Dimensions

Conclusion

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Analyzing the Setup

The Master Equation

The Smart Move

Final Conclusion

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