Answer:
[tex]3.26198\times 10^{-12}\ kg[/tex]
Explanation:
E = Electric field = 20 MN/C
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of drop
The electrical force will balance the weight
[tex]Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg[/tex]
The mass that can be suspended is [tex]3.26198\times 10^{-12}\ kg[/tex]
From charge to mass ratio, the mass of the charges is 3.2 × 10^-12 Kg.
The charge to mass ratio experiment was used by R. A. Millikan to accurately determine the charge to mass ratio of the electron. We have the following information from the question;
Field strength = 20 MN/C
Number of charges = 10
Now;
The magnitude of electric field strength is obtained from;
E = F/q
F = Eq
Where;
F = electric force
E = electric field intensity
q = magnitude of charge
F = 10 × 20 × 10^6 × 1.6 × 10^-19 = 3.2 × 10^-11 N
Where the charges fall freely under gravity;
F = mg
m = F/g
m = 3.2 × 10^-11 N/10 ms-2
m = 3.2 × 10^-12 Kg
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