A long, solid, conducting cylinder has a radius of 2.0 cm. The electric field at the surface of the cylinder is 155 N/C, directed radially outward. Let A, B, and C be points that are 1.0 cm, 2.0 cm, and 4.5 cm, respectively, from the central axis of the cylinder.

What are (a) the magnitude of the electric field at C and the electric potential differences (b) VB-VC and (c) VA - VB?

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mithless670 mithless670 · Answer 1 · 2020-02-28T07:25:46+01:00

Explanation:

The electric field intensity on the surface of cylinder E₁ = λ/2πε₀R I

here λ is the charge density of cylinder

R is the radius of cylinder

and 1/2πε₀ is a constant of permitiivity

Similarly the intensity at a distance r from center of cylinder

E₂ = λ/2πε₀r II

Dividing II by I , we have

E₂/E₁ = R/r

here E₁ = 155 N/C

R = 2.0 cm and r = 4.5 cm

Substituting , the value

E₂/155 = 2/4.5

or E₂ = [tex]\frac{2}{4.5}[/tex] x 155 = 68.9 N/C

( b ) Because E = - dV/dr

Here V is the potential at a point

Thus V = - [tex]\int\limits^a_b {E} \, dr[/tex]

Here E = λ/2πε₀r and b = 2cm and a = 4.5 cm

Thus VB - VC = λ/2πε₀ [ log4.5 - log2 ]

( c ) The potential at points A and B is the same .

Thus VA - VB = 0