Question

a long wire of linear charge density lambda passes through a cube of side l in such a manner that flux through it is maximum. now the position of wire is changed in such a manner that flux is minimum . the ratio of maximum flux to minimum flux is???

Answer 1

Final Answer : √3 : 1

Steps:
1) See Pic.

Answer 2

The ratio of maximum flux to minimum flux is $\sqrt{L^{2}+ B^{2}+ H^{2} } : H$

Explanation:

Given: A long wire of linear charge density lambda passes through a cube of side

To find:  The ratio of maximum flux to minimum flux

Formula used:  $Q = \frac{q}{e_{0} }$

Solution:

A long wire of linear charge density lambda passes through a cube of side l in such a manner that flux through it is maximum.

 $Q = \frac{q}{e_{0} }$

Maximum flux occurs diagonally,

$Q_{max} = \frac{\sqrt{L^{2}+ B^{2}+ H^{2} } }{E_{0} }$

Minimum flux occurs when height equals length

$Q_{min} = \frac{H}{E_{0} }$

The ratio of maximum by minimum flux is given by

$\frac{Q_{max} }{Q_{min} } =\frac{\frac{\sqrt{L^{2} +B^{2}+ H^{2} } }{E_{0} } }{\frac{H}{E_{0} } }$

$\frac{Q_{max} }{Q_{min} } = \frac{\sqrt{L^{2}+ B^{2}+ H^{2} } }{H}$

Final answer:

The ratio of maximum flux to minimum flux is $\sqrt{L^{2}+ B^{2}+ H^{2} } : H$

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