By Squadron Leader Zahid Mir CS&IT Department , Superior University PHY-AP -06 Electric Field due to Line of Charge By Squadron Leader Zahid Mir CS&IT Department , Superior University

Line of Charge Consider a +ve charge Q is distributed x-axis +++++++++++ 2a P qo + r x y-axis z-axis Ѳ dy dQ y Consider a +ve charge Q is distributed uniformly along a line with length ‘2a’ lying along y-axis between y= -a and y=+a. We want to calculate electric field at a point ‘P’ on the x-axis at a distance ‘x’ from the origin. We divide the line into infinitesimal segments each of which acts as a point charge ’dQ’. Let the length of a typical segment at height ‘y’ be ‘dy’. Linear Charge Density  = Q = Total Charge 2a Total Length

Line of Charge Charge density for the segment ‘dy’ is: dQ =  dy dQ = (Q/2a) dy The magnitude of electric field ‘dE’ at point ‘P’ dE = ¼o dQ/r2 dE = ¼o Q dy/ 2a(x2 + y2) Electric field in terms of its x- and y-components dEx = dE Cos & dEy = dE Sin Here Cos = x/r = x /(x2 + y2)1/2 Sin = y/r = y/(x2 + y2)1/2

Line of Charge (cont) Symmetry Argument dEy2 dE2 Ѳ Ѳ O X-axis Ѳ Ѳ dE1 r dEy1 y2 Symmetry Argument For every charge element ‘dQ’ located at position +y1, there is another charge element ‘dQ’ located at –y2. When we add the electric fields due to the charge element at +y1 & -y2 we find the y-components have equal magnitudes but opposite directions; So their sum is zero. dEy = 0 Hence total contribution of electric field is along x-component.

Line of Charge (cont) dEx = dE Cos Here dE = ¼o Q dy/ 2a(x2 + y2) Cos = x/r = x /(x2 + y2)1/2 dEx = ¼o Q dy/ 2a(x2 + y2) {x /(x2 + y2)1/2} Ex =  x dy 4o (x2 + y2)3/2 here tan  = y/x Ex =  1 2o x(1+ x2/a2)1/2 Ex =  2o x a -a

The field magnitude depends only on the distance of point ‘P’ from the line of charge. 2o r E α 1/r The direction of E is radially outward from the line if  is +ve and radially inward if  is –ve.