SATHEE: electric-charges-and-fields Question 50

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Question: Q. 9. Given a uniformly charged plane/sheet of surface charge density $σ = 2 \times 10^{17} C / {cm}^{2}$ .

(i) Find the electric field intensity at a point $A, 5 mm$ away from the sheet on the left side.

(ii) Given a straight line with three points $X, Y$ and $Z$ placed $50 cm$ 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 $A, E = \frac{σ}{2 ε_{0}} = \frac{2 \times 10^{17}}{2 \times 8.85 \times 10^{- 12}}$

$\begin{matrix} (1) & = 1.12 \times 10^{28} N / C \end{matrix}$

(ii) At point $Y$ , Because at $50 cm$ , 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

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