Understanding Charge Distribution on Charged Spheres in Contact

When two charged metallic spheres are brought into contact, charge flows between them until they reach the same electric potential. Since they are conductors, charge can move freely on their surface. Once they are separated, they retain the final charge distribution achieved when they were in contact.

Electric Potential of a Charged Sphere

The electric potential $V$ on the surface of a charged metallic sphere with charge $Q$ and radius $R$ is given by the formula:

$$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} = k \frac{Q}{R}$$

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

Applying Equal Potential Condition

Let the two spheres have radii $R_1$ and $R_2$. When they are in contact and reach equilibrium, their final charges will be $Q_1$ and $Q_2$ respectively. At this point, the electric potential on the surface of both spheres must be equal.

So, $V_1 = V_2$.

Using the formula for potential:

$$k \frac{Q_1}{R_1} = k \frac{Q_2}{R_2}$$

The constant $k$ cancels out from both sides:

$$\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$$

Calculating the Ratio of Final Charges

We are asked for the ratio of final charges $Q_1$ and $Q_2$. From the equation $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$, we can rearrange it to find the ratio $\frac{Q_2}{Q_1}$.

Divide both sides by $Q_1$ and multiply both sides by $R_2$:

$$\frac{Q_1}{R_1 Q_1} \times R_2 = \frac{Q_2}{R_2 Q_1} \times R_2$$

$$\frac{R_2}{R_1} = \frac{Q_2}{Q_1}$$

So, the ratio of the final charges $Q_2$ and $Q_1$ is equal to the ratio of their radii $R_2$ and $R_1$.

$$\frac{Q_2}{Q_1} = \frac{R_2}{R_1}$$

Analysis of Options

Let's compare our derived ratio with the given options:

• Option 1: $Q_2 / Q_1 = R_1 / R_2$. This is incorrect.
• Option 2: $Q_2 / Q_1 < R_1 / R_2$. This is incorrect.
• Option 3: $Q_2 / Q_1 > R_1 / R_2$. This is incorrect.
• Option 4: $Q_2 / Q_1 = R_2 / R_1$. This matches our derived result.

Final Answer Derivation Summary

When two conducting spheres touch, their potentials become equal. Using the potential formula $V = kQ/R$, we set $V_1 = V_2$ for the two spheres with final charges $Q_1$, $Q_2$ and radii $R_1$, $R_2$. This leads to $kQ_1/R_1 = kQ_2/R_2$, simplifying to $Q_1/R_1 = Q_2/R_2$. Rearranging this gives the ratio $Q_2/Q_1 = R_2/R_1$.

Revision Table: Charged Spheres in Contact

Additional Information: Charge Sharing and Capacitance

This phenomenon is related to the concept of capacitance. The capacitance of an isolated spherical conductor with radius $R$ is given by $C = 4\pi\epsilon_0 R = R/k$. The potential is $V = Q/C$. When two conductors are in contact, their potentials are equal, so $V_1 = V_2$. This means $Q_1/C_1 = Q_2/C_2$. Substituting the capacitance values, $Q_1/(R_1/k) = Q_2/(R_2/k)$, which simplifies to $kQ_1/R_1 = kQ_2/R_2$, leading back to $Q_1/R_1 = Q_2/R_2$. Thus, the final charge on each sphere is proportional to its radius (and its capacitance).

If the total initial charge was $Q_{total}$, then $Q_1 + Q_2 = Q_{total}$. Since $Q_1/R_1 = Q_2/R_2$, we have $Q_1 = Q_2 \frac{R_1}{R_2}$. Substituting this into the total charge equation: $Q_2 \frac{R_1}{R_2} + Q_2 = Q_{total}$. This gives $Q_2 ( \frac{R_1}{R_2} + 1) = Q_{total}$, or $Q_2 \frac{R_1+R_2}{R_2} = Q_{total}$. So, $Q_2 = Q_{total} \frac{R_2}{R_1+R_2}$. Similarly, $Q_1 = Q_{total} \frac{R_1}{R_1+R_2}$. This shows how the total charge is distributed based on the radii.