An object is lifted from the surface of aspherical planet to an altitude equal to the radius of the planet.As a result, what hap
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The electric potential V(z) on the z-axis is : V =
The magnitude of the electric field on the z axis is : E = kб 2( 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 ----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv =
V =
=
Therefore the electric potential V(z) =
Also
The magnitude of the electric field on the z axis is : E = kб 2( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
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