The electric potential V(z) on the z-axis is : V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Given data :
V(z) =2kQ / a²(v(a² + z²) ) -z
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex] ----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]
V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]
= [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]
Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]
Also
The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
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