A uniformly charged solid sphere of radius R has potential V0 (measured with respect to infinity) on

Q: Why is the potential at the center $1.5 V_0$?

The potential inside a solid sphere is given by $V = \frac{KQ}{2R^3}(3R^2 - r^2)$. At the center, $r = 0$, so the formula simplifies to $V = \frac{3KQ}{2R}$, which is exactly $1.5$ times the surface potential $V_0 = \frac{KQ}{R}$.

Q: How do we know if a point is inside or outside the sphere?

The potential is maximum at the center and decreases as you move outwards. On the surface, it is $V_0$. Therefore, any potential greater than $V_0$ must correspond to a point inside the sphere, and any potential less than $V_0$ must correspond to a point outside.

Q: Why does the potential decrease outside the sphere?

Outside the sphere, it behaves like a point charge located at the center. The potential is given by $V = \frac{KQ}{r}$, which is inversely proportional to the distance $r$. As $r$ increases, $V$ decreases.

Q: Can the radius $R_1$ be zero?

Yes! A radius of zero simply means the point is located exactly at the center of the sphere. In this problem, the potential $\frac{3V_0}{2}$ corresponds exactly to the center.

Q: What if the sphere was a hollow conducting shell?

If it were a hollow conducting shell, the potential inside would be constant and equal to the surface potential $V_0$. In that case, there would be no point with a potential of $1.5 V_0$ or $1.25 V_0$.

Question

A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere, the equipotential surfaces with potentials $\frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4}$ and $\frac{V_0}{4}$ have radius $R_1, R_2, R_3$, and $R_4$ respectively. Then, Options: (A) $R_1 = 0$ and $R_2 > (R_4 - R_3)$ (B) $R_1 eq 0$ and $(R_2 - R_1) > (R_4 - R_3)$ (C) $R_1 = 0$ and $R_2 < (R_4 - R_3)$ (D) $2R < R_4$

Sigma Percentile · Accepted Answer

Imagine you are an explorer, and your mission is to map the invisible landscape of electric potential surrounding a uniformly charged solid sphere. This problem takes us on a thrilling journey from the very core of the sphere to the vast expanse outside it.

### The Anatomy of a Charged Sphere

Before we dive into the specific radii, we must arm ourselves with the fundamental laws governing the electric potential of a solid sphere. Let the sphere have a radius $R$ and a total charge $Q$. The potential exactly on its surface is our reference point, given by $V_0 = \frac{KQ}{R}$.

However, the potential is not constant everywhere. As we drill into the sphere, the potential increases, reaching its absolute maximum at the dead center. The formula for the potential inside the sphere ($r \leq R$) is a beautiful quadratic curve:

$$V_{in} = \frac{KQ}{2R^3}(3R^2 - r^2) = \frac{V_0}{2R^2}(3R^2 - r^2)$$

Conversely, as we fly away from the sphere, it behaves just like a point charge. The potential outside ($r \geq R$) drops off inversely with distance:

$$V_{out} = \frac{KQ}{r} = V_0\frac{R}{r}$$

With these tools in hand, let's hunt down the mysterious radii $R_1, R_2, R_3$, and $R_4$.

### Decoding the Inner Depths

Our first target is $R_1$, where the potential is $\frac{3V_0}{2}$, or $1.5 V_0$. Notice something special? If we plug $r = 0$ into our inside formula, we get exactly $1.5 V_0$. This means the potential is $1.5 V_0$ right at the center of the sphere! Therefore, without any heavy lifting, we know that **$R_1 = 0$**.

Next, we seek $R_2$, where the potential is $\frac{5V_0}{4}$, or $1.25 V_0$. Since $1.25 V_0$ is greater than the surface potential $V_0$, this point must also lie buried inside the sphere. We set up our equation using the inside formula:

$$\frac{5V_0}{4} = \frac{V_0}{2R^2}(3R^2 - R_2^2)$$

By canceling $V_0$ and rearranging the terms, we get:

$$\frac{5}{2} = 3 - \left(\frac{R_2}{R}\right)^2$$

$$\left(\frac{R_2}{R}\right)^2 = 3 - 2.5 = 0.5$$

Taking the square root, we find **$R_2 = \frac{R}{\sqrt{2}}$**.

### Venturing Beyond the Surface

Now we shift our focus to $R_3$, where the potential drops to $\frac{3V_0}{4}$, or $0.75 V_0$. Because this is less than $V_0$, we have officially left the sphere. We must now use the outside formula:

$$\frac{3V_0}{4} = V_0\frac{R}{R_3}$$

Solving for $R_3$ is straightforward. We simply invert the fraction to get **$R_3 = \frac{4R}{3}$**.

Similarly, for $R_4$, the potential is a mere $\frac{V_0}{4}$. This is far outside the sphere. Using the same logic:

$$\frac{V_0}{4} = V_0\frac{R}{R_4}$$

This immediately yields **$R_4 = 4R$**.

### The Final Showdown: Comparing the Radii

We have successfully mapped all four coordinates:
- $R_1 = 0$
- $R_2 = \frac{R}{\sqrt{2}} \approx 0.707 R$
- $R_3 = \frac{4R}{3} \approx 1.333 R$
- $R_4 = 4R$

Now, let's put the options to the test. We need to evaluate the expression $(R_4 - R_3)$:

$$R_4 - R_3 = 4R - \frac{4R}{3} = \frac{12R - 4R}{3} = \frac{8R}{3} \approx 2.67 R$$

Let's check option (c): Is $R_2 < (R_4 - R_3)$? Yes, $0.707 R$ is definitely less than $2.67 R$. Since we also know $R_1 = 0$, **Option (c) is absolutely correct**.

Let's check option (d): Is $2R < R_4$? Since $R_4 = 4R$, the statement $2R < 4R$ is trivially true. Thus, **Option (d) is also correct**.

By systematically applying the boundary conditions of the electric potential, we transformed a complex spatial problem into a clean, algebraic victory!

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Visualized Solution

$V (r)$ for Solid Sphere

Finding $R_{1}$

Finding $R_{2}$

Calculating $R_{2}$

Finding $R_{3}$

Calculating $R_{3}$

Calculating $R_{4}$

Checking Options

Conclusion

The Sigma Insight: Electric Potential and Potential Difference

The Anatomy of a Charged Sphere

Decoding the Inner Depths

Venturing Beyond the Surface

The Final Showdown

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V(r) for Solid Sphere

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R4​