Calculating Potential Difference in a Uniform Electric Field

[um0123]

Homework Statement

A force of $$4.30\cdot10^{-2}$$Newtons is needed to move a charge of 56 microCoulombs a distance of 20 cm in the direction of a uniform electric field. What is the potentioal difference that will provide this force

Homework Equations

Im trying to find one that would work

The Attempt at a Solution

i have converted 56 microCoulombs into $$5.6\cdot10^{-5}$$ Coulombs and 20 centimeters into .2 Meters. but that's the easy stuff. I can't seem to find an equation that would help me solve this.

 

[ApexOfDE]

Can you write eqs that show relationship between F-E and E-V?

 

[um0123]

i only know a few equations:

$$\Delta~V = k_C\frac{q}{r}$$

$$\Delta~V =-E\Delta~d$$

$$\Delta~V =\frac{PE_{electric}}{q}$$

But i don't see any of these helping, unless I am overlooking something huge.

 

[um0123]

i have one equation for force, but it requires two stationary charged particles, all i have is one.

 

[Cherry_pie]

Hi there,

First, try calculating the E-field, and then use the answer you get to then calculate the p.d. :)

 

[um0123]

the only equation i have to calculate E-Field is:

$$E = \frac{k_Cq}{r^2}$$

and i can't use that because the 56 microcoulomb particle isn't the thing creating the uniform field, it is being acted upon by another field. all i know is that it was moved 20 centimeters with a force of 4.3x10^-2 Newtons.

 

[Cherry_pie]

You should also have the equation F=qE...you can rearrange this to find the electric field, because you know the size of the force that had to act on the charged particle. This field magnitude can then be used to find the potential difference :)

 

[um0123]

oh i forgot about {tex]E=\frac{F}{q_0}[/tex]

solving for E:

$$E=\frac{4.3\cdot10^-2}{5.6\cdot10^-5}$$

$$E= 767.85$$

Then i plug that into $$\Delta~V =E\Delta~d$$

and get the answer 153.57

and i checked in the answers section of my lab manual and it is correct.

Thanks so much for your help!