Question: Q. 9. Given a uniformly charged plane/sheet of surface charge density $\sigma=2 \times 10^{17} \mathrm{C} / \mathrm{cm}^{2}$.
(i) Find the electric field intensity at a point $A, 5 \mathrm{~mm}$ away from the sheet on the left side.
(ii) Given a straight line with three points $X, Y$ and $Z$ placed $50 \mathrm{~cm}$ away from the charged sheet on the right side. At which of these points, the field due to the sheet remains the same as that of point $A$ and why?
A&E [CBSE SQP 2016]
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Solution:
Ans. (i) At point $A, E=\frac{\sigma}{2 \varepsilon_{0}}=\frac{2 \times 10^{17}}{2 \times 8.85 \times 10^{-12}}$
$$ \begin{equation*} =1.12 \times 10^{28} \mathrm{~N} / \mathrm{C} \tag{1} \end{equation*} $$
(ii) At point $Y$, Because at $50 \mathrm{~cm}$, the charge sheet acts as a finite sheet and thus the magnitude remains same towards the middle region of the plane sheet.
[CBSE Marking Scheme, 2016] 1½
