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Two point charges \mathrm{+16 q } and  \mathrm{-4 q }  located at \mathrm{ x=0} and \mathrm{ x=L} . The location of a point on the x-axes from \mathrm{ x=0}, at which the net electric field due to there two charges is zero is -

Option: 1

\mathrm{ l }


Option: 2

\mathrm{ 2 l}


Option: 3

\mathrm{ L / 2}


Option: 4

\mathrm{L / 6}


Answers (1)

best_answer

\mathrm{\frac{k \cdot 16 q}{(l+x)^2} =\frac{4 k q}{x^2}}
\mathrm{\frac{x}{l+x} =\frac{1}{2} }
\mathrm{2 x =l+x }

\mathrm{\\ l =x }

Posted by

Ritika Kankaria

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