Two particles A and B , possessing opposite charges of +2.0 10⁻⁶ C and -2.0 10⁻⁶ C , are positioned at a separation of 1.0 cm . Evaluate the electric field at a point located on the perpendicular bisector of the dipole, 1.0 m away from the centre.

Step-by-step solution

Given charges q = 2.0 10⁻⁶ C and separation 2a = 1.0 cm = 10⁻² m . Dipole moment p = q 2a = 2.0 10⁻⁶ 10⁻² = 2.0 10⁻⁸ C m . Distance from the centre r = 1.0 m . Since r 2a , the electric field at a point on the perpendicular bisector (equatorial plane) is given by E = 1 4 ₀ p r^3 . Substituting the values: E = 9 10^9 2.0 10⁻⁸ (1.0)^3 E = 180 N C ⁻¹ .