Calculating Acceleration Change: Oil Drop in Electric Field | F=ma and F_e=qE

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To calculate the change in acceleration of an oil drop losing an electron in an electric field of 1.00x10^6 N/C, the relevant equations are F=ma and F_e=qE. The initial charge of the oil droplet is assumed to be zero, as no other information is provided, which simplifies the calculation. The charge q used for the change in acceleration should be the charge of one electron, 1.60x10^-19 C. The electric field remains constant despite the loss of the electron, allowing for a straightforward calculation of the final acceleration. The change in acceleration can be determined by applying these principles.
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If an oil drop with mass 1.00x10^-14 kg loses an electron whilst in an electric field of 1.00x10^6 N/C, what is the change in acceleration?

***

F=ma
F_e = qE

So I know delta a for change in acceleration will be (qE/m) final - (qE/m) initial... but what do I use for q (I can't use 1.60*10^-19 C, right, because that's just for one electron and not the whole oil droplet?). Also, will the electric field stay the same?
 
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Would the oil droplet initially be considered neutral, so that the initial component (with Q=0) cancels out?
 
It looks like you have to assume that the initial charge Q0 = 0 since you're not given any other info.

The electric field applied should stay the same; the loss of an electron does not affect the applied field strength.
 

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