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## Homework Statement

A metallic hollow cylinder has a diameter of ##4.2 cm##. Along his axis there is a wire having a diameter of ##2.68 \mu m##(considering it as a hollow cylinder). Between the cylinder and the wire there is a voltage of ##855 V##.

What is the electric field on the wire surface and the cylinder surface?

## Homework Equations

Electric potential:

##\int_{A}^{B} \vec E_0 \cdot d\vec l = V_0(A) - V_0(B)##

For a generic point ##P##

##V_0(x, y, z) = \int_A^P \vec E_0 \cdot d\vec l + V_0(A)##

## The Attempt at a Solution

First I got the radius from the diameters, so:

##D_1 = 2.68\mu m = 2.68 * 10^-6 m##

##D_2 = 4.2 cm = 4.2 * 10^-2 m##

##R_1 = 1.34 * 10^-6 m##

##R_2 = 2.1 * 10^-2 m##

At this point I know that ##V_0(R_1) - V_0(R_2) = 855 V##, so I have to use the first equation to find out ##E_0##. The problem is that I don't understand what ##E_0## I am calculating with this equation:

$$\int_{R_1}^{R_2} \vec E_0 \cdot d\vec l = V_0(R_1) - V_0(R_2)$$

Is it the wire surface? Or am I calculating the electric field inside the cylinder but outside the wire?