- #1
Kernul
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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?