Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric field

Q: Why can we use the formula for a point charge for a metallic sphere?

For points on or outside the surface of a spherically symmetric charge distribution (like a metallic sphere), the electric field and potential behave exactly as if all the charge were concentrated at the center.

Q: Is the potential inside the metallic sphere zero?

No! The electric field inside a conductor is zero, which means the potential is constant throughout the volume and equal to its value on the surface.

Q: What if the spheres were non-conducting and uniformly charged?

The formulas for $E$ and $V$ on the surface would remain exactly the same. The difference only arises for points inside the sphere.

Q: How is surface charge density $\sigma$ related to this?

The electric field just outside a conductor is $E = \sigma/\epsilon_0$. So, the ratio of electric fields $E_1/E_2$ is directly equal to the ratio of their surface charge densities $\sigma_1/\sigma_2$.

Q: Could we solve this by finding the ratio of charges $Q_1/Q_2$ first?

Absolutely! From $E_1/E_2 = R_1/R_2$, you get $(kQ_1/R_1^2) / (kQ_2/R_2^2) = R_1/R_2$, which simplifies to $Q_1/Q_2 = (R_1/R_2)^3$. Substituting this into the potential ratio formula gives the same result.

Question

Consider two charged metallic spheres $S_1$ and $S_2$ of radii $R_1$ and $R_2$, respectively. The electric fields $E_1$ (on $S_1$) and $E_2$ (on $S_2$) on their surfaces are such that $E_1/E_2 = R_1/R_2$. Then the ratio $V_1$ (on $S_1$)/$V_2$ (on $S_2$) of the electrostatic potentials on each sphere is

Options:

(A) $\left(\frac{R_1}{R_2}\right)^3$

(B) $\frac{R_2}{R_1}$

(C) $\frac{R_1}{R_2}$

(D) $\left(\frac{R_1}{R_2}\right)^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine we have two charged metallic spheres, $S_1$ and $S_2$. Their radii are $R_1$ and $R_2$ respectively. We are given a very specific and interesting condition regarding the electric fields on their surfaces: the ratio of their electric fields $E_1/E_2$ is exactly equal to the ratio of their radii $R_1/R_2$.

Our ultimate goal is to find the ratio of the electrostatic potentials on the surfaces of these two spheres, which is $V_1/V_2$.

### The Master Equation

To solve this elegantly, we need to recall the fundamental formulas for the electric field $E$ and the electrostatic potential $V$ on the surface of a charged sphere.

The electric field is given by:
$$E = \frac{kQ}{R^2}$$

And the electrostatic potential is given by:
$$V = \frac{kQ}{R}$$

Look closely at these two equations. There is a beautiful mathematical relationship hiding in plain sight. If we take the expression for the electric field and multiply it by the radius $R$, we get:
$$E \cdot R = \left(\frac{kQ}{R^2}\right) \cdot R = \frac{kQ}{R}$$

This perfectly matches our formula for the potential! Therefore, we can establish a powerful master equation:
$$V = E \cdot R$$

### Final Calculation

Now that we have our master equation, finding the ratio of the potentials is a breeze. We can write the ratio $V_1/V_2$ as:
$$\frac{V_1}{V_2} = \frac{E_1 \cdot R_1}{E_2 \cdot R_2}$$

We can separate this into two distinct ratios:
$$\frac{V_1}{V_2} = \left(\frac{E_1}{E_2}\right) \cdot \left(\frac{R_1}{R_2}\right)$$

The problem generously provides us with the condition that $\frac{E_1}{E_2} = \frac{R_1}{R_2}$. Let's substitute this crucial piece of information into our equation:
$$\frac{V_1}{V_2} = \left(\frac{R_1}{R_2}\right) \cdot \left(\frac{R_1}{R_2}\right)$$

Multiplying these together yields our final, elegant result:
**$$\frac{V_1}{V_2} = \left(\frac{R_1}{R_2}\right)^2$$**

This simple yet profound relationship highlights how interconnected electrostatic properties are on the surface of conductors.

Select Answer:

Visualized Solution

$S_{1}$ and $S_{2}$

Formulas for $E$ and $V$

Relation between $E$ and $V$

Ratio of Potentials

Substituting Given Condition

Final Answer

Food for Thought

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup

The Master Equation

Final Calculation

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