Conducting droplet

Electricity and Magnetism Level pending

A conducting sphere bubble of radius a \large a and thickness t < < a t<<a is charged to a potential V \large V . Now it collapses to form a spherical droplet. The potential of droplet is

( 3 a t ) V \Bigg(\dfrac{3a}{t}\Bigg)V ( 2 a t ) ( V ) 1 2 \Bigg(\dfrac{2a}{t}\Bigg)(V)^{\frac{1}{2}} ( a 3 t ) 1 3 V \Bigg(\dfrac{a}{3t}\Bigg)^{\frac{1}{3}}V ( 2 a t ) 1 3 \Bigg(\dfrac{2a}{t}\Bigg)^{\frac{1}{3}}

Sparsh Sarode by Sparsh Sarode , 22, India

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1 solution

Sparsh Sarode
May 31, 2016

Initial potential, V = K q a V=\dfrac{Kq}{a} where q is the charge on the bubble

Volume=surface area . Thickness

Volume, V = ( 4 π a 2 ) ( t ) V=(4\pi a^2)(t)

This should be equal to the final volume, V ′ V'

V ′ = 4 3 π r 3 = ( 4 π a 2 ) ( t ) V'=\frac{4}{3}\pi r^3=(4\pi a^2)(t)

r = ( 3 a 2 t ) 1 3 r=(3a^2t)^{\frac{1}{3}}

New potential V 1 = K q r V_1=\dfrac{Kq}{r}

V 1 = ( K q ( 3 a 2 t ) 1 3 ) ( a a ) V_1=\Bigg(\dfrac{Kq}{(3a^2t)^{\frac{1}{3}}}\Bigg)\Bigg(\dfrac{a}{a}\Bigg)

V 1 = ( a 3 t ) 1 3 V V_1=\Bigg(\dfrac{a}{3t}\Bigg)^{\frac{1}{3}}V

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