A charge Q is distributed over three concentric spherical shells of radii a, b, c (a

Q: Why is the potential inside the shell equal to the surface potential?

Because the electric field inside a hollow conductor is zero. Since $E = -dV/dr = 0$, the potential $V$ must be constant throughout the interior, matching its value at the surface.

Q: Why don't we use the distance $r$ in the potential formula?

Since $r < a$, the point $P$ is inside all three shells. The potential inside a shell doesn't depend on the distance from the center; it's a constant value everywhere inside.

Q: What if the charge densities were different for each shell?

If the charge densities were different, we would need additional information (like the specific ratio of charges) to find the individual charges $Q_a, Q_b,$ and $Q_c$.

Q: How did we get the area of the shells?

The shells are spherical, and the surface area of a sphere is given by the geometric formula $4\pi R^2$, where $R$ is the radius of the sphere.

Q: Can we use Gauss's Law to find the potential directly?

Gauss's Law is used to find the electric field. To find the potential, you would integrate the electric field. Since Gauss's Law tells us $E=0$ inside the shells, integrating it confirms that the potential is constant.

Question

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ ($a < b < c$) such that their surface charge densities are equal to one another.
The total potential at a point at distance $r$ from their common centre, where $r < a$ would be

Options:

(A) $\frac{Q(a^2+b^2+c^2)}{4\pi\varepsilon_0(a^3+b^3+c^3)}$

(B) $\frac{Q(a+b+c)}{4\pi\varepsilon_0(a^2+b^2+c^2)}$

(C) $\frac{Q}{4\pi\varepsilon_0(a+b+c)}$

(D) $\frac{Q(ab+bc+ca)}{12\pi\varepsilon_0 abc}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the very center of three massive, hollow spheres, nested inside one another like Russian Matryoshka dolls. The innermost shell has a radius $a$, the middle one has a radius $b$, and the outermost has a radius $c$.

We are tasked with finding the electric potential at a point $P$, located at a distance $r$ from the center. The crucial detail here is that $r < a$. This means our point $P$ is buried deep inside all three of these spherical shells!

### The Master Equation for Charge

The problem hands us a beautiful symmetry: the surface charge density ($\sigma$) is identical for all three shells. 
$$\sigma_a = \sigma_b = \sigma_c = \sigma$$

What exactly is surface charge density? It is simply the total charge spread over a surface divided by the area of that surface. For a sphere, the surface area is $4\pi R^2$. Therefore, we can express the charge on each individual shell in terms of this common density $\sigma$:
$$Q_a = \sigma(4\pi a^2)$$
$$Q_b = \sigma(4\pi b^2)$$
$$Q_c = \sigma(4\pi c^2)$$

We are also told that the total charge across all three shells is $Q$. This gives us our conservation equation:
$$Q_a + Q_b + Q_c = Q$$

Substituting our expressions for the individual charges, we get:
$$\sigma(4\pi a^2) + \sigma(4\pi b^2) + \sigma(4\pi c^2) = Q$$

Factoring out the common terms, we can solve for $\sigma$:
$$\sigma = \frac{Q}{4\pi(a^2 + b^2 + c^2)}$$

Now, we can find the exact fraction of the total charge that resides on each shell. For example, the charge on the innermost shell is:
$$Q_a = \frac{a^2}{a^2 + b^2 + c^2}Q$$
The charges $Q_b$ and $Q_c$ follow the exact same pattern.

### The Magic of Potential Inside a Shell

Here is where the physics gets truly elegant. What is the electric potential at a point **inside** a uniformly charged spherical shell?

Because the electric field inside a hollow conductor is zero, it takes absolutely no work to move a test charge around inside it. This means the potential difference is zero, and the potential everywhere inside the shell is constant. It is exactly equal to the potential on its surface!
$$V_{	ext{inside}} = V_{	ext{surface}} = \frac{1}{4\pi\varepsilon_0}\frac{q}{R}$$

Since our point $P$ is at a distance $r < a$, it lies inside all three shells. Therefore, the total potential at $P$ is simply the sum of the surface potentials of each shell:
$$V_P = V_a + V_b + V_c$$
$$V_P = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q_a}{a} + \frac{Q_b}{b} + \frac{Q_c}{c}ight)$$

### Final Calculation

Let's bring it all together. We substitute our expressions for $Q_a$, $Q_b$, and $Q_c$ into the potential equation:
$$V_P = \frac{1}{4\pi\varepsilon_0}\left(\frac{a^2 Q}{a(a^2+b^2+c^2)} + \frac{b^2 Q}{b(a^2+b^2+c^2)} + \frac{c^2 Q}{c(a^2+b^2+c^2)}ight)$$

Notice how the $a$ in the denominator cancels one $a$ in the numerator, and the same happens for $b$ and $c$. Factoring out the common terms, we arrive at our final, elegant result:
$$V_P = \frac{Q}{4\pi\varepsilon_0(a^2+b^2+c^2)}(a + b + c)$$

This matches option (b) perfectly. The beauty of this problem lies in recognizing that the potential inside a shell is constant, turning what looks like a complex calculus problem into a straightforward algebraic sum!

Select Answer:

Visualized Solution

$a < b < c$

$σ_{a} = σ_{b} = σ_{c} = σ$

Charge on Shells

Total Charge $Q$

Individual Charges

Potential Inside a Shell

Total Potential at $P$

Substitution

Final Answer

What if $a < r < b$ ?

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup

The Master Equation for Charge

The Magic of Potential Inside a Shell

Final Calculation

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