Given Info: The length of a straight thinwire is 2m. it is uniformly charged and its charge density becomes 1.5 x 10-6 C/m.To find: the electric FIELD intensity due to the wire at a point 1.5 m away from the centre of the wire is...Solution: charge density of wire, λ = 1.5 x 10-° C/mwe can apply Coulomb's LAW only for point charge, so here we have to first take an elementary length of wire to MAKE it valid for coulomb's law.so CUT an element dx, x distance from the point of observation. [ see figure] now charge on element, dq = Xdx electric field density due to element, dE = KdQ/x²= Kλdx/x2= KXJdx/x²= KX [-1/x]upper LIMIT= (1 + d)lower limit = d= KA [1/d - 1/(1+ d)]= KXI/d(I + d)now putting, K = 9 x 10^9 Nm²/C², λ = 1.5 x 10-6 C/m, d = 0.5 m and I = 2m= (9 x 10^9 x 1.5 x 10^-6 x 2)/(0.5 x 2.5)= 12 x 10³/(1.25)= 9600 N/CTherefore the electric field intensity due to wire at a point 1.5m away from the centre of wire is 9600 N/C.pls mark me as brainliest and follow me