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

Step 1: Write expression for electric potential of a charged bubble.
Electric potential of a charged spherical bubble is given by: \[ V = \frac{kQ}{R} \]
Step 2: Consider charge and radius after coalescence.
Let each small bubble have charge \(q\) and radius \(r\). After coalescence: \[ \text{Total charge } = 3q \] Since volume is conserved, \[ \frac{4}{3}\pi R^3 = 3 \times \frac{4}{3}\pi r^3 \Rightarrow R = 3^{1/3} r \]
Step 3: Find potential of the bigger bubble.
\[ V_{\text{big}} = \frac{k(3q)}{3^{1/3}r} = 3^{2/3}\frac{kq}{r} \]
Step 4: Find the required ratio.
\[ V_{\text{small}} : V_{\text{big}} = \frac{kq}{r} : 3^{2/3}\frac{kq}{r} = 1 : 3^{2/3} \]
Final Answer: \[ \boxed{1 : 3^{2/3}} \]