JEE PHYSICS MCQ: ELECTRIC FIELDS AND CHARGES

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Application +4 | -1

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field

100V\;{m^{ - 1}}.

If the mass of the drop is

1.6 \times {10^{ - 3}}g,

the number of electrons carried by the drop is

\left( {g = 10m{s^{ - 2}}} \right)

View answer and explanation

Correct option: D

{10^{12}}

Option D (

{10^{12}}

) is correct because the number of electrons is determined by equating the electric force to the gravitational force and then dividing the charge by the elementary charge.

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