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In the Milikan oil drop experiment the drop carries 10 electronic charges charge and a mass of 4.4 xx 10^(-15)kg. It is held almost stationary between two horizontal plate separated by 2cm. What is the potential between the plates? |
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Answer» Solution :Data supplied Let V be the potential the plates separated by a distance d. Thus, electric field intensity, `E=V/d, d=2 xx 10^(-2)m` Charge `q=ne= 10 xx 10^(-19)C, Mass m=4.4 xx 10^(-15)kg` Since the drop is held stationary, the electrostatic field is EQUAL to the WEIGHT of the drop. i.e.qE=mg `(q V)/(d)=mg` `V=(mgd)/(q)=(4.4 xx 10^(-15) xx 9.8 xx 2 xx 10^(-2))/(10 xx 1.6 xx 10^(-19))=539"volts"` |
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