Question

6. Consider a line charge with uniform charge density λ lying on the x-axis from x =-L to x = 0. a) Determine the electric field a distance y above the right end of the line charge (point P in the figure) and a distance r to the right of the line charge (point P2 in the figure). Pi C-I b) In lecture you saw the electric field of an infinite line charge. Now we wil consider a "semi-infinite" line charge; that is, a line charge that is infinite on only one side. In other words, suppose we take the left end of the line charge to be infinitely far to the left, but the line charge still stops at r = 0 What is the electric field at P and P2 in this case? c) Use your result from part b) and superposition to derive the electric field of an infinite line charge. (Just worry about point P1 for this part.)

Answer 1

given charge density lambda a) Electric field at point P_1 = E_1 dE_1 = k(lambda dx)/(x^2 + y^2) (x/squareroot x^2 + y^2 i^^+ y/squareroot x^2 + y^2 j^^) Integrating from x = -L to x = 0 E_1 = k lambda integral (xi/(x^2 + y^2)^3/2 + yj^^/(x^2 + y^2)^3/2)dx E_1 = k lambda {[1/squareroot y^2 + l^2 - 1/y]i^^+ l/y squareroot y^2 + l^2 j^^} Electric field at P_2 = E_2 dE_2 = k lambda dx/(x + X)^2 i^^ Integrating E_2 = k lambda integral dx/(x + X)^2 i^^= k lambda l/X(X - l) b) for l rightarrow +infinity, E_1 = k lambda {[-i^^/y + j^^/squareroot 0 + 1]} E_1 = k lambda {-i^^/y + j^^/y} = k lambda/y (-i^^+ j^^) \r\n Similiarly E_2 = -k lambda/X(1) = -k lambda/X c) for Infinite wire the result becomes E_1 = k lambda/y