Deepak Patra
Last Activity: 10 Years ago
Sol. q base 2 = 2 × 10^–6 C, q base 1^2 = – 4 × 10^–6 C, r = 20 cm = 0.2 m
(E base 1 = electric field due to q base 1, E base 2 = electric field due to q base 2)
⇒ (r - x)^2/x^2 = -q base 2/q base 1 ⇒ (r - 1)^2/x = -q^2/q base 1 = 4 * 10^-6/2 * 10^-6 = 1/2
⇒ (r/x - 1) = 1/√2 = 1/1.414 ⇒ r/x = 1.414 + 1 = 2.414
⇒ x = r/2.414 = 20/2.414 = 8.285 cm