c). In the Millikan oil–drop apparatus a drop of radius r = 1.63 µm has an exces

Question

c). In the Millikan oil–drop apparatus a drop of radius r = 1.63 µm has an excess charge of 4 electrons. What are the magnitude of the electric field that is required to balance the drop so it remains stationary in the apparatus?

Use for the density of oil the value ? = 0.925 g/cm³.

Express the result to three significant figures. If you must use scientific notation, please enter as follows: ex. 0.00012 = 1.2E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number (right: -1.00, wrong, - 1.00).

d). Find the separation distance (positive value) between two point charges + 3 e and - 6 e if the magnitude of the electric force F between them is 0.71 x 10-10 N.

Express the result in the unit [nm] and to three significant figures. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number: e.g. right: -1.00, wrong: - 1.00.

Explanation / Answer

(c)

force balance : 

qE = mg ; 

or 4e E =  V g ; 

or   E =  (  V g  ) / ( 4 e ) ; 

V = 4 / 3      ( 1.63 E-6 ) ^3  = 1.715728 E-17 m^3 ; 

= 0.925 g / cm^3 = 925 kg / m^3 ; 

so E =  [ 925 *(  1.715728 E-17  ) * 9.8  ]  / [ 4 * 1.6 E-19 ] ; 

or E = 2.430 E+5 

( d ) 

F = k ( 3e ) ( 6e ) / r^2 ; 

or r^2 = [  k ( 3e ) ( 6e ) ]  / F ; 

or r^2 = [ 9 E+9  *  18 * ( 1.6 E-19 ) ^2 ] / [ 0.71 E-10 ] ;

or r = 7.6427 E-9 ;

or r = 7.6427 nm

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