Question

a stationary Oil Drop between two parallel plates has a charge of 3.2 into 10 to the power minus 19 coulomb and a weight of 1.6 into 10 to the power minus 14 Newton find the electric field acting on the drop
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Answer 1

Answer:The electric field acting on the drop is $E = 5\times10^{4}\ N/C$.

Explanation:

Given that,

Charge $q = 3.2\times10^{-19}$

Weight $mg =1.6\times10^{-14}$

We know that,

$F = qE$....(I)

From newton's law

$F = mg$

Put the value of F in equation (I)

$mg = qE$

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

$E =\dfrac{1.6\times10^{-14}N}{3.2\times10^{-19}C}$

$E = 5\times10^{4}\ N/C$

Hence, The electric field acting on the drop is $E = 5\times10^{4}\ N/C$.