- #1

Kennedy111

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## Homework Statement

A 1.50 x 10^-14 kg oil drop accelerates downward at a rate of 1.80 m/s^2 when placed between two horizontal plates that are 9.40 ch apart. The potential difference between the two plates is 980 V. Determine the magnitude of the charge on the oil drop.

m = 1.50 x 10^-14 kg

a = 1.80 m/s^2 (down)

d = 9.40 cm = 9.40 x 10^-2 m

V = 980 V

## Homework Equations

Fnet = Fg + Fe

Fnet = ma

Fe = Eq

Fg = mg

E = V/d

## The Attempt at a Solution

Fnet = Fg + Fe

Fe = Fnet - Fg

= ma - mg

= ((1.50 x 10^-14 kg)*1.80 m/s^2)) - ((1.50 x 10^-14 kg)(9.81 m/s^2))

= -1.2015 x 10^-13 N

E = V/d

= (980 V) / (9.40 x 10^-2 m)

= 10425.53191 V/m

Fe = Eq

q = Fe / E

= (-1.2015 x 10^-13 N) / (10425.53191 V/m)

= -1.1524592 x 10^-17 C

The charge on the oil drop is approx. -1.15 x 10^-17 C ?