The quarter disc of radius ‘R’ (shown in figure) has a uniform surface charge density 'σ'.

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The quarter disc of radius ‘R’ (shown in figure) has a uniform surface charge density $' σ'$ .
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Electric potential $(0, 0, Z)$ is $\frac{σ}{4 ε_{0}} (\sqrt{R^{2} + Z^{2}} - Z)$

x and y components of electric field at $(0, 0, Z)$ are zero

Electric potential at $(0, 0, Z)$ is $\frac{σ}{8 ε_{0}} (\sqrt{R^{2} + Z^{2}} - Z)$

Z- component of electric field at $(0, 0, Z)$ is $\frac{σ}{8 ε_{0}} (1 - \frac{Z}{\sqrt{R^{2} + Z^{2}}})$

answer is A, B.

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Detailed Solution

We know that electric potential due to entire disc is
$\begin{array}{l} V = \frac{σ}{2 ε_{0}} (\sqrt{R^{2} + Z^{2}} - Z) \\ \Rightarrow V d u e t o q u a r t e r d i s c i s \\ V = \frac{σ}{8 ε_{0}} (\sqrt{R^{2} + Z^{2}} - Z) \\ \Rightarrow E l e c t r i c f i e l d (Z c o m p o n e n t i s) \\ E = - \frac{\partial v}{\partial z} \\ E = - \frac{σ}{8 ε_{0}} (\frac{1}{2 \sqrt{R^{2} + Z^{2}}} (2 Z) - 1) \\ E = - \frac{σ}{8 ε_{0}} (1 - \frac{Z}{\sqrt{R^{2} + Z^{2}}}) \end{array}$