Question

An infinite line of positive charge lies along the y axis, with charge density λ = 2.30 μC/m. A dipole is placed with its center along the x axis at x = 28.0 cm. The dipole consists of two charges ±10.0 μC separated by 2.00 cm. The axis of the dipole makes an angle of 45.0° with the x axis, and the positive charge is farther from the line of charge than the negative charge. Find the net force exerted on the dipole.

Answer 1

(1cm) sint5 = 12 cm Force exexted on point charge q in electric field É is given by F=qE _ x (cm) Sinas"=1 cm As per the Gauss Law, electric field at distance x from infinite wire E = 2 Here x= linear charge density - Go = permittivity of free space = 8.854 xjö? Film 26 EX force exerted on positive char for positive charge x= 280cm + 1 cm = 28.707 cm = 0.29m Electric field at location of positive charge 1 Ex = (2.30x10 to cfm) - 1.434103 N/c 2 (3:14) (8.254 xid*f/m) (0.29m). Ft = (1046) E+ = (10x1õʻc) (143xlo?n\c ) =[1-43N] for negative charge x = 28.0am-12 cm = 27.3 cm = 0.27 m Electric field at location of negative charge E- = (2:30 x1o6c/m) - = 1.53x105 N/c 2C3-14) (8.854x10"//m] (0-27m) – Force exerted on Negative charge F = (-10 dec) ČE-) = (loxíoc)(1-53X10% nie) = 1.53N Net force F = Ft + F_ = 1-43N+(-1.53N) =-O'IN F = 0.IN/ This force will be dierected along megative x direction.