Two particles A and B , possessing charges of +2.00 10⁻⁶ C and -4.00 10⁻⁶ C respectively, are held fixed at a separation of 20.0 cm . Find the point(s) on the line AB at which the electric field is zero.

Step-by-step solution

Let the point where the electric field is zero be at a distance x from charge A . Since the charges have opposite signs, the null point must lie outside the line segment joining the two charges, on the side of the charge with the smaller magnitude. Here, |q_A| At the null point, the magnitudes of the electric fields due to both charges must be equal: 1 4 ₀ |q_A| x^2 = 1 4 ₀ |q_B| (x + d)^2 Substituting the given values: 2.00 10⁻⁶ x^2 = 4.00 10⁻⁶ (x + 20.0)^2 1 x^2 = 2 (x + 20.0)^2 Taking the square root of both sid