(II) An oil drop whose mass is 2.8 x 10-15 Kg is held at rest between two large plates separated by 1.0 cm (Fig. 27-3), when the potential difference between the plates is 340 V. How many excess electrons does this drop have?
Solution 3P: We have to calculate the number of electron possessed by an oil drop in Millikan’s oil drop experiment owing to which it remain suspended in between the air. Step 1 of 3 Concept: Weight of the oil drop is Fg= mg is directed directly downwards due to Earth’s gravitational force. Electric force acting on the charged oil drop is Fe= qE which is directed in the vertically upward direction. Here, q Charge on the oil drop E Electric field between the plates Electric field between the two parallel plates separated by the distance d when a potential difference of V is applied across it ends, is given as, E = V d Step 2 of 3 15 Mass of the oil drop is m = 2.8 × 10 kg Separation between the plates is d = 1.0 cm = 0.01 m Potential difference between the plates is V = 340 V 19 Charge of an electron is e = 1.6 × 10 C 2 Acceleration due to gravity is g = 9.80m/s As the oil is at rest in equilibrium state,therefore the gravitational force in the downward direction on the oil drop must be equal to the upward electric force on the same oil drop, so F g F e mg = qE V mg = q (d) q = mgd V (2.8×101k) (9.8 m/s ) (0.01 m) q = (340 V) 19 q = 8.07 × 10 C 19 Thus, the magnitude of excess charge on the electron is q = 8.07 × 10 C .