Question

Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:

• A. \(1 : 2^{2/3}\)
• B. \(1 : 3^{1/3}\)
• C. \(1 : 3^{2/3}\)
• D. \(3^{2/3} : 1\)

Hint:

When identical charged spheres coalesce, charge adds linearly but radius changes according to volume conservation.

Answer 1

The Correct Option is C

To solve this problem, let's analyze the situation of three smaller bubbles coalescing to form a bigger bubble. The problem involves understanding the physics of charged spherical conductors, namely the relationship between charge, radius, and electric potential.  

1. The potential \( V \) on a charged spherical bubble is given by the formula: \(V = \frac{kQ}{R}\), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the bubble.
2. When three bubbles coalesce, the total volume remains constant. This can be expressed as: \(3 \cdot \frac{4}{3}\pi r_0^3 = \frac{4}{3}\pi R^3\) Simplifying this, we get: \(3r_0^3 = R^3\) Thus, the radius of the larger bubble is: \(R = r_0 \cdot 3^{1/3}\)
3. Since all three initial bubbles have identical charge \( q \), the total charge on the larger bubble is: \(Q_{total} = 3q\)
4. The potential of one of the smaller bubbles is: \(V_{small} = \frac{kq}{r_0}\)
5. The potential of the larger bubble is: \(V_{big} = \frac{k \cdot 3q}{R} = \frac{k \cdot 3q}{r_0 \cdot 3^{1/3}}\) Simplifying, we have: \(V_{big} = \frac{3^{2/3} \cdot kq}{r_0}\)
6. The ratio of potentials between the smaller bubble and the larger bubble is: \(\frac{V_{small}}{V_{big}} = \frac{\frac{kq}{r_0}}{\frac{3^{2/3} \cdot kq}{r_0}} = \frac{1}{3^{2/3}}\)

Therefore, the ratio of the potentials on one initial bubble to that on the resultant bigger bubble is \(1 : 3^{2/3}\).