Answer to Question #200756 in Mechanics | Relativity for Rocky Valmores

Question

The Millikan Oil Drop Experiment calculated the charge of the electron using the concept of velocity dependent force. When electric field is applied, the upward motion of the particle is given by qE – mg – bv = F where mg is the weight of the charged particle, bv is the retarding force, and qE is the force on the charge due in an electric field. a. Derive the expression for the terminal velocity of the oil drop as a function of time using F= -mg - bv.

Answer 1

Solution.

$F=-mg-bv;$

$ma=-mg-bv;$

$a=\dfrac{v-v_0}t; v_0=0;$

$m\dfrac{v}{t}=-mg-bv;$

$mv=-mgt-bvt;$

$v(m+bt)=-mgt;$

$v=\dfrac{-mgt}{m+bt};$

Answer:$v=\dfrac{-mgt}{m+bt};$