Question

Two wires of linear charge density lamda passing through a sphere of radius R and a cube of side R so that the flux linked with them is maximum. Then the ratio of flux of sphere to the cube is ___________

Answer 1

Given:

Two wires of linear charge density lamda passing through a sphere of radius R and a cube of side R so that the flux linked with them is maximum.

To find:

Ratio of flux of sphere to cube ?

Calculation:

First of all, let's consider that the linear charge density of the wire is $\lambda$.

• Now, in case of sphere, the max flux will be when the wire will be along the diameter (so as to hold max charge).

• Similarly, max flux in cube will be when the wire will be along the body diagonal (so as to hold max charge).

$\therefore \: q_{sphere} = \lambda \times l$

$\implies \: q_{sphere} = \lambda \times (2r)$

______________________________________

$\therefore \: q_{cube} = \lambda \times l$

$\implies\: q_{cube} = \lambda \times ( \sqrt{3} r)$

Now, applying GAUSS' LAW:

$\therefore \: \dfrac{ \phi_{sphere}}{ \phi_{cube}} = \dfrac{ \bigg \{ \dfrac{ \lambda(2r)}{ \epsilon_{0}} \bigg \} }{ \bigg \{ \dfrac{ \lambda( \sqrt{3}l) }{ \epsilon_{0} } \bigg \}}$

$\implies \: \dfrac{ \phi_{sphere}}{ \phi_{cube}} = \dfrac{2}{ \sqrt{3} }$

So , the ratio is 2 : √3 .