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  • A thin infinite sheet charge and an infinite line charge of respective charge densities <img alt="+\sigma \text { and }+\lambda" src="https://entrancecorner.oncodecogs.com/gif.latex?&plus;

A thin infinite sheet charge and an infinite line charge of respective charge densities +\sigma \text { and }+\lambda are placed parallel at 5 m distance from each other. Points 'P' and 'Q' are at \frac{3}{\pi} \mathrm{m} \text { and } \frac{4}{\pi} \mathrm{m} perpendicular distances from line charge towards sheet charge, respectively. 'EP' and 'EQ' are the magnitudes of resultant electric field intensities at points 'P' and 'Q' respectively. If \frac{E_p}{E_Q}=\frac{4}{a} \text { for } 2|\sigma|=|\lambda| then the value of a is _________

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \mathrm{E}_{\mathrm{p}}=\frac{\sigma}{2 \varepsilon_0}-\frac{2 \mathrm{~K} \lambda}{\mathrm{r}} \\ & \mathrm{E}_{\mathrm{p}}=\frac{\sigma}{2 \varepsilon_0}-\frac{\lambda}{2 \pi \varepsilon_0\left(\frac{3}{\pi}\right)} \\ & \mathrm{E}_{\mathrm{p}}=\frac{2 \sigma}{2 \varepsilon_0}-\frac{2 \sigma}{6 \varepsilon_0}=\frac{\sigma}{6 \varepsilon_0} \end{aligned}

Similarly,  \mathrm{E}_{\mathrm{Q}}=\frac{\sigma}{2 \varepsilon_0}-\frac{2 \sigma}{2 \pi \varepsilon_0\left(\frac{4}{\pi}\right)}=\frac{\sigma}{4 \varepsilon_0}

\begin{aligned} & \frac{E_p}{E_Q}=\frac{4}{6} \\ & a=6 \end{aligned}

Posted by

Deependra Verma

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