Question

a linear charge havig linear charge density lamda penetrates a cube diagonally and then it penetrate a sphere diametrically what will be the ratio of flux coming out of cube and sphere

Answer 1

Hiii.....

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“If a linear charge having a linear charge density penetrates a cube diagonally, and then it penetrates a sphere diametrically… “

Did you mean a line of charge with a uniform charge density passes through the diagonally opposite corners of a cubic Gaussian surface and also along the diameter of a spherical Gaussian surface …

If you did, and if the cube diagonal is L and the diameter of the sphere is D, and the charge per unit length of the line of charge is lambda, the flux through the surface of the cube would be proportional to L * lambda, and the flux through the surface of the sphere would be proportional to D * lambda. Provided that neither of the two surfaces contains any other charges.

In that case the ratio would of course be L/D.

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Hope it will help u......

Answer 2

Given:

A linear charge

Linear charge density $\[ = \lambda \]$

To Find: Ratio of flux coming out of cube and sphere

Solution:

Consider a cube of side '$a$' through which a linear charge penetrates diagonally as shown in the figure below.

Linear charge density is given by the formula:

$\[\lambda = \frac{Q}{L}\]$

where, $Q=$ charge and $L=$ length

In the given cube, the length of the diagonal can be calculated by the Pythagoras theorem.

$\[\begin{gathered} d = \sqrt {2{a^2} + {a^2}} \hfill \\ d = \sqrt {3{a^2}} \hfill \\ \end{gathered} \]$

Therefore, for the cube, the linear charge is given as:

$\[\[{Q_1}\] = \lambda \times \sqrt {3{a^2}} \]$                ... (i)

Now, consider a sphere of radius '$a$' through which a linear charge penetrates diametrically as shown in the figure below.

Therefore, the diameter of the sphere is

$d=2a$

Thus, for the sphere, the linear charge is given as:

$\[\[{Q_2}\] = \lambda \times 2\alpha \]$                     ... (ii)

Since flux is given as:

$\[{\phi _E} = \frac{{{Q_{net}}}}{{{\varepsilon _o}}}\]$

Therefore, the ratio of the flux coming out of cube and sphere can be calculated as:

$\[\begin{gathered} \frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{{Q_1}/{\varepsilon _o}}}{{{Q_2}/{\varepsilon _o}}} \hfill \\ \frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{\lambda \times \sqrt 3 a}}{{\lambda \times 2a}} \hfill \\ \frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{\sqrt 3 }}{2} \hfill \\ \end{gathered} \]$

Hence, the ratio of the flux coming out of cube and sphere is $\[\frac{{\sqrt 3 }}{2}\]$.

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