Question

A charge q is placed inside a sphere of radius ‘a’ filled with water and another charge 2q is placed inside a cube of side ‘2a’ which is vacuumed inside. Find the ratio of the flux linked with the sphere to that linked with the cube. (Take relative permittivity of water as 80.)

Answer 1

From Gauss’s theorem:

∅ = `q/(epsilon_r epsilon_o)`    ...[Where ε_r is relative permittivity of medium inside Gaussian surface.]

Flux through the sphere (water inside):

Water has relative permittivity:

ε_r = 80

`phi_"sphere" = q/(80 epsilon_o`

Flux through the cube (vacuum inside):

`phi_"cube" = (2q)/epsilon_o`

Now take the ratio:

`phi_"sphere"/phi_"cube" = (q/(80 epsilon_o))/((2q)/epsilon_o)`

= `q/(80 epsilon_o) xx epsilon_o/(2q)`

Cancel q and ε_o:

= `1/(80 xx 2)`

= `1/160`