Electrostatics 7 Question 26

28. A conducting bubble of radius $a$, thickness $t(t \ll a)$ has potential $V$. Now the bubble collapses into a droplet. Find the potential of the droplet.

$(2005,2 \mathrm{M})$

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Answer:

Correct Answer: 28. $V^{\prime}=V \Big(\frac{a}{3t}\Big)^{1 / 3}$

Solution:

  1. Let $q$ be the charge on the bubble, then

$$ \begin{aligned} V & =\frac{K q}{a} & \text { Here, } K=\frac{1}{4 \pi \varepsilon_{0}} \\ \therefore \quad q & =\frac{V a}{K} & \end{aligned} $$

Let after collapsing, the radius of droplet becomes $R$, then equating the volume, we have

$$ \left(4 \pi a^{2}\right) t=\frac{4}{3} \pi R^{3} $$

$$ \therefore \quad R=\left(3 a^{2} t\right)^{1 / 3} $$

Now, potential of droplet will be $V^{\prime}=\frac{K q}{R}$

Substituting the values, we have

$$ V^{\prime}=\frac{(K) \frac{V a}{K}}{\left(3 a^{2} t\right)^{1 / 3}} \quad \text { or } \quad V^{\prime}=V \Big(\frac{a}{3t}\Big)^{1 / 3} $$



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