Calculating Potential of Combined Spherical Drops

This problem involves the principles of conservation of volume and conservation of electric charge when multiple identical spherical drops of a liquid combine to form a single large drop. We need to determine the potential of this large drop given the potential of the small drops.

Step-by-Step Solution

Let's break down the process:

1. Determine the relationship between the radius of the large drop and the radius of a small drop using the conservation of volume.
2. Determine the relationship between the total charge on the large drop and the charge on a single small drop using the conservation of charge.
3. Relate the potential of a spherical drop to its charge and radius.
4. Use the relationships from steps 1 and 2, and the given potential of the small drop, to calculate the potential of the large drop.

Detailed Calculation of Drop Potential

Let:

• $r$ be the radius of each small spherical drop.
• $q$ be the electric charge on each small spherical drop.
• $V_{small}$ be the potential of each small spherical drop.
• $R$ be the radius of the single large spherical drop.
• $Q$ be the total electric charge on the large spherical drop.
• $V_{large}$ be the potential of the large spherical drop.

We are given that the potential of each small drop is $V_{small} = 9V$. The potential of a spherical drop is given by the formula:

$\qquad V = \frac{1}{4\pi\epsilon_0} \frac{Q'}{r'}$

where $Q'$ is the charge and $r'$ is the radius. Let $k = \frac{1}{4\pi\epsilon_0}$ be the electrostatic constant.

So, for a small drop:

$\qquad V_{small} = \frac{kq}{r} = 9V \quad (*)$

Conservation of Volume

When 8 identical small drops combine, their total volume is conserved and forms the volume of the large drop. The volume of a sphere is $\frac{4}{3}\pi r^3$.

Total volume of 8 small drops = $8 \times \left(\frac{4}{3}\pi r^3\right)$

Volume of the large drop = $\frac{4}{3}\pi R^3$

By conservation of volume:

$\qquad \frac{4}{3}\pi R^3 = 8 \times \left(\frac{4}{3}\pi r^3\right)$

Cancel out $\frac{4}{3}\pi$ from both sides:

$\qquad R^3 = 8 r^3$

Taking the cube root of both sides:

$\qquad R = (8r^3)^{1/3} = 2r \quad (**)$

The radius of the large drop is twice the radius of a small drop.

Conservation of Charge

The total charge is also conserved when the drops combine. The total charge on the 8 small drops is the sum of their individual charges.

Total charge on 8 small drops = $8 \times q$

Charge on the large drop = $Q$

By conservation of charge:

$\qquad Q = 8q \quad (***)$

The charge on the large drop is eight times the charge on a small drop.

Potential of the Large Drop

Now we can find the potential of the large drop using its charge $Q$ and radius $R$:

$\qquad V_{large} = \frac{kQ}{R}$

Substitute the relationships from $(**)$ and $(***)$ into this equation:

$\qquad V_{large} = \frac{k(8q)}{(2r)}$

Simplify the expression:

$\qquad V_{large} = \frac{8kq}{2r} = 4 \left(\frac{kq}{r}\right)$

From equation $(*)$, we know that $\frac{kq}{r} = 9V$. Substitute this value into the expression for $V_{large}$:

$\qquad V_{large} = 4 \times (9V)$

$\qquad V_{large} = 36V$

Thus, the potential of the single large drop formed by combining eight identical spherical drops, each having a potential of 9V, is 36V.

Summary Table

Conclusion on Combined Potential

The potential of the large drop is found to be 36V, which is four times the potential of each small drop. This result comes directly from applying the principles of conservation of volume and charge, and the formula for the potential of a charged sphere.

Revision Table: Key Concepts for Potential of Combined Drops

Additional Information: Electrostatic Potential and Capacitance

The concept of electric potential is fundamental in electrostatics. Potential difference drives the flow of charge (current). For a charged conductor like a spherical drop, all points on its surface are at the same potential. Inside a conductor, the electric field is zero, and the potential is constant and equal to the surface potential.

Combining charged drops is also related to capacitance. The capacitance of a sphere is given by $C = 4\pi\epsilon_0 r$. The potential is related to charge and capacitance by $V = Q/C$.

For the small drop: $V_{small} = \frac{q}{C_{small}}$, where $C_{small} = 4\pi\epsilon_0 r$.

For the large drop: $V_{large} = \frac{Q}{C_{large}}$, where $C_{large} = 4\pi\epsilon_0 R$.

Using $Q = 8q$ and $R = 2r$:

$C_{large} = 4\pi\epsilon_0 (2r) = 2 (4\pi\epsilon_0 r) = 2C_{small}$.

$V_{large} = \frac{8q}{2C_{small}} = 4 \frac{q}{C_{small}} = 4 V_{small}$.

This confirms the result obtained earlier. The potential increases because although the charge increases, the radius (and thus capacitance) also increases, but the charge increases faster relative to the radius in the potential formula $V=Q/R$ (since $Q \propto N$, $R \propto N^{1/3}$, $V \propto N/N^{1/3} = N^{2/3}$). Here $N=8$, $N^{2/3} = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4$.