Answer to Question #97490 in Mechanics | Relativity for Ak

Question #97490
In an experiment, the tiny droplets of the negatively electrified oil are dropping through a vac-
uum. Magnitude of the electric field is 5.92 × 104 N C−1
. Direction of the electric field is assumed to be
downward. (a) An experimenter observes a one particular droplet and that droplet remains suspended against
gravity. Assume the mass of the droplet to be 2.93 × 10−15 kg. What is the charge carried by the droplet? (b)
Now another droplet of same mass falls, in the vacuum, from rest to 15.3 cm in 0.275 s. What is the charge
on the droplet in this case?
1
Expert's answer
2019-10-29T10:53:47-0400

(a) Let q1q_1 be charge on droplet

Gravitational force due to weight == electrostatic force due to electric field

So,


q1E=mgq_1E=mg


q1=mgE=2.93×1015×9.85.92×104=4.85×1019 Cq_1=\frac{mg}{E}=\frac{2.93\times 10^{-15}\times9.8}{5.92\times 10^4}=4.85\times 10^{-19}\ C

(b) Let q2q_2 be charge on this droplet

then according to question,


s=ut+12at2.153=0.5a×(.2752)ora=4.05 m sec2s=ut+\frac{1}{2}at^2\\.153=0.5a\times(.275^2) \\or\\a=4.05\ m\ sec^{-2}

balancing force,


ma=mgq2Ema=mg-q_2E

m=2.93×1015 kgE=5.92×104m=2.93\times 10^{-15}\ kg\\E=5.92\times 10^{4}


2.93×1015×4.05=2.93×1015×9.8q2×5.92×1042.93\times 10^{-15}\times 4.05=2.93\times 10^{-15}\times 9.8-q_2\times 5.92\times 10^4

Orq2=2.84×1019 COr\\q_2=2.84\times 10^{-19} \ C


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