Practice portal › Electric Charge and Properties › Motion of a Charge in an Electric Field
Asked in GSEB Board July 2015; GSEB Board March 2019 · Balanced drop and Millikan
Given: a drop of mass M and charge q held at rest in a field E.
Idea: the drop is balanced when the upward electric force equals its weight.
qE = Mg
E = (Mg)/q
The field must point so that the force on the drop is upward (upward for a positive drop, downward for a negative one).
So the field needed is (Mg)/q.
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