Question

WOT the Find the electric potential V at a distance z (at point P-se of a uniform line charge the equation 30 (chapter 2) of the textbook. see figure below) above the center 6. distribution (length 2L, charge density You might want to use z I 2L

Answer 1

line change density = lambda. let us first consider an element of the distribution dx at x change in the element dq, dq = lambda dx now the potential at p due to dq, dv = 1/4 pi epsilon_0 dq/squareroot z^2 + x^2 here the net potential at p due to the complete distribution v(z) = integrals dv = 1/4 pi epsilon_0 integral^L_x = -L lambda dx/squareroot z^2 + x^2 = lambda/4 pi epsilon_0 2 integral^L_0 dx/squareroot x^2 + z^2 or v(z) = 2. lambda/4 pi epsilon_0 [ln Vertical-bar x squareroot x^2 + z^2 Vertical-bar ]^L_0 or, v(z) = 2 lambda/4 pi epsilon_0 [ln (L + squareroot L^2 + z^2) - ln z] or, v(z) = 2 lambda/4 pi epsilon_0 ln (L + squareroot L^2 + z^2/z).