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Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward.a) An electric field is established by applying a potential difference to the plates. It is found that a field of strength E0 will cause the droplet to be suspended motionless. Write an expression for the droplet's charge q. Let g be the acceleration due to gravity. Express your answer in terms of the suspending field E0 and the droplet's weight mg.

Respuesta :

Answer:

[tex]q=\frac{mg}{E_o}[/tex]              

Explanation:

Given:

Charge = q

Electric field strength =[tex]E_o[/tex]

weight of the droplet = mg

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, [tex]F=qE_o[/tex]

This is balanced by the weight, [tex]mg[/tex]

Equating the two:

[tex]qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}[/tex]

The expression for the droplet's charge q should be considered as the [tex]q = mg / Eo[/tex].

Calculation of the expression:

Since

Charge = q

Electric field strength = Eo

weight of the droplet = mg

The charge should be considered as the suspended motionless. This is due to the the electric force on the charge is balanced by the weight of the droplet.

Now

electric force on charged droplet, should be F - qEo

This is balanced by the weight, i.e.. mg

So, The expression for the droplet's charge q should be considered as the [tex]q = mg / Eo.[/tex]

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