09. A negatively charged oil drop is prevented from failing under grav

1.	09. A negatively charged oil drop is prevented from failing under gravity byapplying a vertical electric field 100 V m-1. If the mass of the drop is 1.6 x 10g, the number of electrons carried by the drop is (g = 10 ms)a) 10^18b) 10^6c) 10^9d) 10^12
Answer» Answer: 10^(19) electrons (if mass of DROP given is correct) Explanation: WEIGHT of the drop = mg Electric force = Eq Both the forces are acting in EQUAL & opposite directions to prevent FALLING of drop Therefore, mg = Eq q = mg / E = (1.6 × 10 g × 10 m/s²) / (100 V/m) = 1.6 C Number of electrons = q/e = (1.6 C) / (1.6 × 10^(-19)) = 10^(19) P.S: Please CHECK the mass of drop which you've provided, there must be a typo in it.

09. A negatively charged oil drop is prevented from failing under gravity byapplying a vertical electric field 100 V m-1. If the mass of the drop is 1.6 x 10g, the number of electrons carried by the drop is (g = 10 ms)a) 10^18b) 10^6c) 10^9d) 10^12

Answer»

Answer:

10^(19) electrons (if mass of DROP given is correct)

Explanation:

WEIGHT of the drop = mg

Electric force = Eq

Both the forces are acting in EQUAL & opposite directions to prevent FALLING of drop

Therefore,

mg = Eq

q = mg / E

= (1.6 × 10 g × 10 m/s²) / (100 V/m)

= 1.6 C

Number of electrons = q/e

= (1.6 C) / (1.6 × 10^(-19))

= 10^(19)

P.S: Please CHECK the mass of drop which you've provided, there must be a typo in it.

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09. A negatively charged oil drop is prevented from failing under gravity byapplying a vertical electric field 100 V m-1. If the mass of the drop is 1.6 x 10g, the number of electrons carried by the drop is (g = 10 ms)a) 10^18b) 10^6c) 10^9d) 10^12​

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09. A negatively charged oil drop is prevented from failing under gravity byapplying a vertical electric field 100 V m-1. If the mass of the drop is 1.6 x 10g, the number of electrons carried by the drop is (g = 10 ms)a) 10^18b) 10^6c) 10^9d) 10^12