Two masses M 1 and M 2 carry positive charges Q 1 and Q 2 , respectively. They are dropped to the floor in a laboratory set up from the same height, where there is a constant electric field vertically upwards. M 1 hits the floor before M 2 . Then,

Step-by-step solution

Time of fall = 2 h a ret Net acceleration of charged masses is a net = g - q E m As, M 1 hits the floor before M 2 ⇒ 2 h a 1 > 2 h a 2 ⇒ 1 a 1 > 1 a 2 ⇒ a 2 > a 1 When reciprocal is taken in equality sign is reversed, then. g - Q 1 E M 1 > g - Q 2 E M 2 ⇒ - Q 1 E M 1 > - Q 2 E M 2 ⇒ Q 1 E M 1 Q 2 E M 2 Here, multiplication with - 1reverse sign of inequality. So, Q 1 M 1 Q 2 M 2 or M 2 Q 1 M 1 Q 2 ⇒ M 1 Q 2 > M 2 Q 1