A very long cylinder of radius 2.60 cm carries a uniform charge density of 2.50

Question

A very long cylinder of radius 2.60 cm carries a uniform charge density of 2.50 nC/m .
part A: Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of 12.0 V .
Part B: Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of 24.0 V .
Part C:Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of 36.0 V .

Explanation / Answer

Va - vb = E * dr

Lower limit: a

Upper limit: b

= 2.50 nC/m

I have solved, using Gauss' Law, the electric field of this cylinder.

E = /(20)(r)

Now to integrate, I removed the /(20) factor out of the integrand and did this:

Vr - V.026 = /(20) * (1/r)dr

Lower limit for this integral: r (the distance from the axis of the cylinder)

Upper limit for this integral: 2.60 cm (the radius of the cylinder)

So Vr - V.026 = Vr =(/(20) * ln(.026/r)

If V.026 = 12

Vr = ((2.5 * 10-9) /(2**8.85*10-12)) * ln(0.026/x) + 12

Vr = 3.3954 cm

If V.026 = 24

Vr = ((2.5 * 10-9) /(2**8.85*10-12)) * ln(0.026/x) + 24

Vr = 4.43 cm

If V.026 = 36

Vr = ((2.5 * 10-9) /(2**8.85*10-12)) * ln(0.026/x) + 36

Vr = 5.79 cm

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