Potential difference and Kinetic energy

1. Mar 6, 2006

donjt81

A 4.0g object carries a charge of 20uC. The object is accelerated from rest through a potential difference, and afterward the ball is moving at 2.0 m/s. What is the magnitude of the potential difference?

I know i have to use the law of conservation

Potential_initial + KE_initial = Potential_final + KE final
Potential_initial + 0 = 0 + KE final

but I dont know what to do after this. can some one give me any pointers?

2. Mar 6, 2006

chroot

Staff Emeritus
The object gained some amount of kinetic energy, which can found with

$KE = \frac{1} {2}mv^2$

This energy had to come from somewhere; in this case, it came from an electric field.

The amount of energy gained by an object of charge q, moving through a potential difference of V volts is:

$E = qV$

The easiest way to realize this is simply to look at the units. The "volt" is exactly equal to one joule per coulomb. Look at how the factors cancel if you use this definition of the volt:

${\text{20}}\,\mu {\text{C}} \cdot V\frac{{{\text{joules}}}} {{{\text{coulomb}}}}$

The coulombs cancel, leaving you with just joules.

All you need do is to solve for V, such that the gain in kinetic energy is equal to the loss of electrical potentail energy:

$\frac{1} {2}mv^2 = qV$

- Warren