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Three infinite plane sheets carrying uniform charge densities $-\sigma, 2 \sigma, 4 \sigma$ are placed parallel to $X Z$ plane at $Y=a, 3 a$, 4a respectively. The electric field at the point $(0,2 a, 0)$ is
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The correct answer is:
$-\frac{7 \sigma}{2 \varepsilon_{0}} \hat{j}$

$\begin{aligned} \mathrm{E}_{\text {net }} &=7 \mathrm{E} \\ &=7 \frac{\sigma}{2 \varepsilon_{0}} \text { alone }-\text { ve } \mathrm{Y} \text {-axis } \\ & \therefore \overrightarrow{\mathrm{E}}=-\frac{7 \sigma}{2 \varepsilon_{0}} \hat{\mathrm{j}} \end{aligned}$
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