Determine the magnitude of the electric field

[William Bush]

Homework Statement

P and Q are points within a uniform electric field that are separated by a distance of 0.1 meters as shown. The potential difference between P and Q is 50 V. Determine the magnitude fo this electric field.

Homework Equations

E=F/q
E=(k)(q)/r^2

The Attempt at a Solution

This problem doesn't fit the pattern of the other "electric field" problems that I have worked. The formulas above are for electric field, and electric field of a point charge; neither one fits this application because I'm not dealing with point charges so there is no value for "q" and I'm not given enough info to find "F". I'm told that the 50 V is the potential difference between P and Q but how does that info fit into my problem?

 

[rootX]

What's the relationship between force and work?

 

[William Bush]

Thanks for the quick response!

1. Work = (F)(d); but how does this help? Should I consider the consider the 50 V between points P and Q as the force?

2. When I do so, I come up with:
E = (8.99 x 10^9)(50 x 10^-6)/0.1^2

When I plug that into my calculator I get 44950000 or 4.495 x 10^7. My soloution sheet says the answer should be 500. Neither of the numbers I got should be rounded up that far. Am I missing something?

 

[rootX]

Thanks for the quick response!

1. Work = (F)(d); but how does this help?

yep, you will find that out.

So, now you know what's F equal to, and d is given.

and assuming that initially the energy between two charges was 0, what would the work equal to?

 

[William Bush]

I'm sorry,...but I'm not following. Isn't the work simply equal to 50 V x .1 meters?

 

[rootX]

no, rather think about potential energy, and the way it is related to the work,
and then find a relationship between potential energy and voltage
and you just found the relationship between Electric field and force,and
between force and the work.

 

[William Bush]

Okay, I got it!...my mistake was that I was trying to use E=kq/r^2. When I plugged the numbers into E=F/q it came out correct. Big thanks to rootX for the help!

 

[rootX]

so, here's the derivation in proper way:
U=qV
F=qE
Fd=U

hence V=Ed

 

[William Bush]

I spoke to soon...I didn't come out correctly my way! Hope you can bear with my so that I can see if I follow your last post

 

[rootX]

yep, i knew

 

[William Bush]

What does "U" stand for in your equations?

 

[rootX]

potential energy

as work = potential energy-0

 

[William Bush]

I would have never been able to come up with the derivations that you listed! I didn't know that work is equal to potential energy -0; or that Fd = U. F = qE is in my textbook so I was aware of that one. I'm worried because I don't understand how you came up with those derivations.

 

[rootX]

I assumed that intially the points are at infinite distance from each other. (a common assumption)
and, then their potential energy changed to U (some value as distance between them is finite).

and work = change in potential energy

and as work = Fd

so U = Fd
U = q* voltage
and qV=qEd