Question

A bubble of conducting liquid is charged to potential
V. It has a radius R and thickness ∆t ( ∆t<< R). It
collapses to form a droplet. The potential at the
surface of droplet is​

Answer 1

Answer:

ANSWER

Potential at the surface of bubble , V=

a

kq

or q=

k

Va

Since bubble collapses into droplet of radius R. So (4πa

2

)t=

3

4

πR

3

or R=(3a

2

t)

1/3

Now the potential of the droplet is V

′

=

R

kq

=

(3a

2

t)

1/3

Va

=V(

3t

a

)

1/3

烘培 i他河流怕有