Question #8e958

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Answer 1 · 2017-04-25T16:35:45

For parallel plates, we know that electric field strength #E# is:

#E = (Delta V)/(Delta x) = 98 * 10^3 " N/C"#

Balancing forces in the vertical direction:

#E q = mg#

#q = (mg)/E = (10^(-6) * 9.81)/(98 * 10^3) approx 10^(-10) \ C#

Answer 2 · 2017-04-25T16:37:35

#10^-10 C#

Since the oildrop is at equilibrium, downward weight = upward electric force.

#F_G=F_E#
#mg=qE#
#mg=qV/d#
#(1*10^(-6))(9.81)=q(980)/(1*10^(-2))#
#q=1.00*10^(-10)C#