# Electric potential, current question

1. Mar 21, 2004

### dura

QUESTION: a small dust mote of mass $$3.2x10^-12$$kg is between two metal plates (one + one -)which can be conected to a 20V power supply. When the power supply is turned on, an electric field of 400 V/m is established. It is observed that the mote starts from rest when the field is turned on (t=0) and travels 1.0 mm in 10s.

a) what is the charge on the mote?
b) the mote moves toward the positive plate, what is the sign of the charge?
c) what is the direction of the electric field between the plates?

Alright... I have a really hard time with these questions, and frankly dont know where to start or how to tackle this. Can someone please help me out with this?

2. Mar 21, 2004

### dura

sorry, my math text writing sucks. The mass should read 3.2x10 (superscript) -12

3. Mar 21, 2004

### Chen

(b) is easiest so start there. If the mote moves towards the positive plate, it means it is attracted to positive charges. What kind of charge is attracted to another positive charge? Negative of course.

(a) Because the electric field is constant, so is the electric force that pulls the mote. If the force on the mote is constant, so is its acceleration.
$$a = \frac{F_{ele}}{m} = \frac{Eq_m}{m}$$.
The mote's initial velocity is 0, so its displacement is given by:
$$x = \frac{1}{2}at^2$$
Rescuing the acceleration from there you get to:
$$a = \frac{Eq_m}{m} = \frac{2x}{t^2}$$
And that's easy to solve.

(c) Since $$\vec F_{ele} = \vec Eq$$, and as we saw the mote's charge is negative, the electric field is in the opposite direction of the electric force. And you already know the direction of the force since you know the mote is attracted to the positive plate. (The direction of the field can be explained in a number of other ways as well, but in our case this explanation is the most straightforward.)