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## Homework Statement

Show that the electric field needed to make the rise time of the oil drop equal to the its field free time is

ε = (2mg)/q

## Homework Equations

Newton's second law

**F**= m

**a**

The force of gravity

**F**

_{G}= mg, where g = 9.8 ms

^{-2}

The force of buoyancy F

_{b}= b

**v**

The force of an electric field on a charge ε =

**F**/q, where ε = the strength of the electric field and q = the charge of the particle

## The Attempt at a Solution

First, I added the forces for the oil drop falling when the electric field is "off." I added the buoyant force that results from terminal velocity and the force of gravity.

**F**= m

**a**= 0 = bv - mg = 0

**[1]**

This can be rearranged to

bv = mg

**[2]**

Second, I reasoned that when the electric field is "on," the sum of the forces must be

**F**= m

**a**= εq - mg = mg

**[3]**

I rewrote this as

ε = (2mg)/q

Was my approach correct for solving this problem?