Kevin Nash
Last Activity: 11 Years ago
Sol. m = 1 g = 10^–3 kg, u = 0, q = 2.5 × 10^–4 C ; E = 1.2 × 10^4 N/c ; S = 40 cm = 4 × 10^–1 m
a) F = qE = 2.5 × 10^–4 × 1.2 × 10^4 = 3 N
So, a = F/m = 1/10^-3 = 3 * 10^3
E base q = mg = 10^-3 * 9.8 = 9.8 * 10^-3 N
b) S = 1/2 at^2 or t = √2a/g = √2 * 4 *10^-1/3 * 10^3 = 1.63 * 10^-2 sec
v^2 = u^2 + 2as = 0 + 2 * 3 * 10^3 * 4 * 10^-1 = 24 * 10^2 ⇒ v = √24 * 10^2 = 4.9 * 10 = 49 m/sec
work done by the electric force w = F→td = 3 * 4 * 10^-1 = 12 * 10^-1 = 1.2 J