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In milky an experiment an Oil Drop of radius .0.000010cm apart if the Drop has charge of 5e over it calculate the potential difference between the plates the density of iron be taken as 1.5

Alok shaw , 6 Years ago
Grade 12
anser 1 Answers
Eshan

Last Activity: 6 Years ago

Dear student,

In the experiment, the oil drop with excess charges on it is kept in the air due to electric force on it by keeping between plates with a potential difference between them.

Under equilibrium,

mg=qE
\implies \rho(\dfrac{4}{3}\pi r^3)g=q\dfrac{V}{d}
\implies V=\dfrac{4\rho \pi r^3 dg}{3q}

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