Question

Determine the intensity of electric field which can balance a drop of oil mass of 0.001 mg carrying 3 electronic charge

Answer 1

Electric filed will be equal to $E=2.041\times 10^{13}N/C$

Explanation:

We have given mass of rain drop m = 0.001 mg = $0.001\times 10^{-3}kg=10^{-6}kg$

Acceleration due to gravity $g=9.8m/sec^2$

So force due to gravity on the rain drop W = mg, here m is mass and g is acceleration due to gravity

So W = $10^{-6}\times 9.8=9.8\times 10^{-6}N$

Charge is given three electronic charge

So charge $Q=3e=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C$

Electric force on the rain drop is given as F = qE, here q is charge and E is electric field , this electric force will balance the weight of the drop

So $qE=9.8\times 10^{-6}$

$E\times 4.8\times 10^{-19}=9.8\times 10^{-6}$

$E=2.041\times 10^{13}N/C$

So electric filed will be equal to $E=2.041\times 10^{13}N/C$

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Answer 2

golmaal hai bhai sab golmaal