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An oil drop with 10 excess electrons is held stationary under a constant electric field of 3 xx 10^(4) NC^(-1) in Millikan oil drop experiment. The density of the oil is 1.26 xx 10^(3) kg m^(-3). Calculate the radius of the drop. |
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Answer» Solution :Data SUPPLIED, r=radius of the droplet, d=density of oil drop `=1.26 xx 10^(3) kg m^(-3)` `g=9.8 m//s^(2), ` n=no. of excess electrons =10 `e=1.6 xx 10^(-19)C""E=3 xx 10^(4) NC^(-1)` As the oil drop is STATIONARY, weight of the droplet=force due to ELECTRIC field `i.e 4/3 pir^(3) dg=ne E, r^(3) =(3ne E)/(4PI dg) =(3 xx 10 xx 1.6 xx 10^(-19) xx 3 xx 10^(4))/(4 xx 3.14 xx 1.26 xx 10^(3) xx 9.8)= 0.9285 xx 10^(-18)` `r=(0.9285 xx 10^(-18))^(1/3)=0.9756 xx 10^(-6)=9.76 xx 10^(-7) m` |
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