If the lines are drawn in proportion to the charge, what is the ratio $\frac{q_{1}}{q_{2}}$ ?

$\frac{3}{8}$

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$3$

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$\frac{1}{6}$

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cannot be said

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Solution

The correct option is A
$\frac{3}{8}$

since, the relative number of EFL originating from or terminating on a charge is proportional to the magnitude of the charge. This ratio of their charges $\frac{q_{1}}{q_{2}}$ must be equal to the ratio of number of electric field lines emanating from $q_{1} a n d q_{2} .$

By carefully while observing the picture, $\frac{q_{1}}{q_{2}} = \frac{6}{16} = \frac{3}{8}$

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