A very long insulating cylinder of charge of radius 2.60 cm carries a uniform linear density of 15.0 nC/m . Part A If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 200 V

Given Information:  

Radius = ra = 2.60 cm = 0.026 m

Density = J = 15.0 nC/m

change in potential difference = ΔV = 200 V

Required Information:  

Distance = d = ?

distance = 0.088 m

Explanation:

As we know

ΔV = Vb - Va = J/4πε₀*ln(rb/ra)

Where ra and rb is the point where potential difference is Va and Vb respectively

1/4πε₀ = 9x10⁹ N.m²/C²

We want to find the distance d = rb - ra

ΔV = J/4πε₀*ln(rb/ra)

200 = 9x10⁹*15x10⁻⁹*ln(rb/ra)

200/135 = ln(rb/ra)

1.48 = ln(rb/ra)

taking e on both sides yields

e^(1.48) = rb/ra

4.39 = rb/ra

rb = 4.39*0.026

rb = 0.114 m

Therefore, the required distance is

d = rb - ra

d = 0.114 - 0.026

d = 0.088 m

Therefore, the other probe must be placed 0.088 m from the surface so that the voltmeter reads 200 V

the components of a vector are the parts of the vectors that are to each other.

b perpendicular