Question

Calculate the electric field strength required to just support an oil drop of mass

Answer 1

Answer:

$E= 3.06 \times 10^{-17} N/C$

Explanation:

As we know that the oil drop experiment tells us the balancing the gravitational force by electrostatic force.

So ,

Electrostatic force = Gravitational force

qE=mg

Here , q is charge , E is electric field , m is mass and g is gravity.

so,

$E=\frac{mg}{q}$

Insert values from question,

$E=\frac{10^{-2}kg \times 9.8 m/s^{2}}{3.2 \times 10^{-19}C}$

or

$E= 3.06 \times 10^{-17} N/C$

This is the required value of electric field.