two point charges of 2 micro Coulomb but opposite in sign are placed 10 cm apart calculate the electric field at a point distant 10 cm from the mid point on the axial line of the dipole Share with your friends Share 18 Vikas answered this as given in question that their are two charges of 2×10-6 coloumb so these charges act as a dipole distance of seperation d=(2a) = 10 cm=10-1mdistance from mid point on axial line r = 10 cmsince p =q×2a = 2×10-6 ×10 -1=2×10-7as we know that the electric feild due to dipole on axial line E = 14πε2pr2E= 9×109 2×2×10-710-2E=36×104 N/C ans.. 22 View Full Answer Shruthy Del answered this .hello rahul, first of all the eqn 4:the gn qtn is E=2*2aq/r^(3) i.e rcube also gn , r=10cm, 2a=10cm, q=2*10^(-6) substituting evrytingin the gn eqn v get E=2*10*2*10^(-6)/10^(3) =40*10^(-6-3) =4*10^(-8) hope my ans. is right...... 1 Shruthy Del answered this yep sorry I forgot to multiply k in the eqn actually the ans I guess shd b 360N/C dry 4 the inconvenience. -8
