Calculating Potential Difference in a Uniform Electric Field

[exitwound]

Homework Statement

A force of 0.053 N is required to move a charge of 37 mC a distance of 25 cm in an electric field. What is the size of the potential difference between the two points?

Homework Equations

$$V_f-V_i=\int \vec{E} \cdot d\vec{S}$$
$$E=\frac{F}{q}$$

The Attempt at a Solution

$$V_f-V_i=\int \vec{E} \cdot d\vec{S}$$
$$\Delta V=Ed$$
$$\Delta V=\frac{Fd}{q}$$
$$\Delta V=\frac{(.053)(.25)}{37e^-3}= .358 V$$

However, the answer sheet says 360V. Am I missing something stupid? I assume it's a constant electric field and the particle is being moved along the field lines.

 

[kuruman]

Are you sure it is 37 mC and not 37 μC?

37 mC = 37x10^-3 C

37 μC = 37x10^-6 C

The second choice gives an answer that is 1000 times the answer of the first choice. Just the factor you need.

 

[exitwound]

Yeah. I copied/pasted it directly from the homework sheet. It says "mC" which I took as 10^-3.

So I was on the right track though...

 

[kuruman]

The only other thing I can think of is whether you got the units of the answer right. Is it perhaps 360 mV? If it is actually 360 V, my last suggestion is that you ask the person who assigned this problem to you for clarification. I cannot find fault with your solution.

 

[exitwound]

Well, it's for a class that's online that I'm not in. Just Googling for practice problems.

http://www.farraguttn.com/science/milligan/physics/ElecAsgn.htm

Question #19 - Answers at the bottom.

 

[kuruman]

I see. Sometimes wrong answers are posted. I have seen physics textbooks with as many as 14% incorrect answers in the back of the book.

Since you are doing this for practice, rest assured that your solution is correct and that you know how to handle this type of problem, so move on.