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Nesterboy [21]
2 years ago
12

A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

e xy plane.
What is the electric potential V(z) on the z axis as a function of z , for z>0 ?
What is the magnitude E of the electric field on the z axis, as a function of z , for z>0 ?
Physics
1 answer:
sleet_krkn [62]2 years ago
7 0

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

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  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

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