An electric dipole of length 8 cm consists of two equal and opposite charges +q and -q . The electric potential at a point on the axial line, which is closer to the positive charge, is given as 8 9 Kq 10^2 V (where K = 1 4 ₀ ). The distance of this point from the centre of the dipole is:

Step-by-step solution

Let the distance of the point from the centre of the dipole be r . The exact formula for the electric potential on the axial line of a dipole (closer to +q ) is: V = K(q 2a) r^2 - a^2 Given length of the dipole, 2a = 8 cm a = 4 cm . Converting distances to metres, 2a = 8 10⁻² m and r^2 - a^2 = (r^2 - 16) 10⁻⁴ m ^2 (where r is in cm). Substituting the given values: V = K q 8 10⁻² (r^2 - 16) 10⁻⁴ = 800Kq r^2 - 16 We are given that V = 8 9 Kq 10^2 = 800Kq 9 . Equating the two expressions: 800Kq r^2 - 16 = 800Kq 9 r^2