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Two parallel line charges +\lambda and -\lambda are placed with a separation R in free space. The net electric field exactly mid way between the two line charge is -

Option: 1

zero


Option: 2

\frac{2 \lambda}{\pi \epsilon_0 K}


Option: 3

\frac{\lambda}{\pi \epsilon_0 R}


Option: 4

\frac{\lambda}{4 \pi \epsilon \nu R}


Answers (1)

best_answer

 we know that,
\mathrm{E=\frac{2 k \lambda}{R} }
but,
\mathrm{\begin{aligned} & E_{\text {Net }}=\left|E_1\right|+\left|E_2\right| \\ &=2\left(\frac{2 k \lambda}{R}\right)=\frac{4 \lambda}{4 \pi t 0 R} \\ &=\frac{\lambda}{\pi t_0 R} \end{aligned} }

Posted by

Ajit Kumar Dubey

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