# Quick clarification/question on Electric Field at a point

1. Feb 28, 2017

### Pablo1122

1. The problem statement, all variables and given/known data
http://i.imgur.com/wBdUomo.png

Another simpler way to ask the question is if the electrical potential at a point is zero, is the electric field at that point also zero?

2. Relevant equations
v = kq/r
E=kq/r^2
F=qe

3. The attempt at a solution

http://i.imgur.com/wBdUomo.png

Alright, so I've done the part a) in the following manner.

http://i.imgur.com/jB9oOP0.png

Now, this is the first step in part b)

http://i.imgur.com/BOKG70A.png

We solved for cos theta by using adjacent/hypotenuse from the triangle we made earlier.
Now, assume the question said that the distance is way more smaller than x. d <<<<<x.

For the first question, that would make it so everything is over 50 which makes electric potential at point P zero. However, in part b what would we replace cos theta with? I was thinking of doing 0/50 (since d is virtually zero) and then the E1 is equal to zero. I'm not sure if that is what should be done

Last edited by a moderator: Feb 28, 2017
2. Feb 28, 2017

### mjc123

I think part a is right, and I think part b is a factor of 1000 out, though I can't tell what you've done. Please show your full working if you want any more help.

3. Feb 28, 2017

### Pablo1122

All I need is in regards to part b. I'm sorry if I was disrespectful I just don't know how to go about writing it. This is the scenario I'm stuck with.

So i'm given a horizontal distance d which is d = 1cm. And I get a vertical distance which is x = 50 cm. Now we we can create a right traingle. Now if we did adjacent/hypotenuse. (Half of the d over x) We'd get 0.5/50.

However, let's assume all that's given is that d is much smaller than x. d <<<< x.

Would I write 0/50 which is then 0? This part multiplies with an equation so if I did it this way the entire equation would be zero.

4. Feb 28, 2017

### kuruman

The way to handle this is to find an algebraic expression for what you need first. Add any fractions that need adding. Then factor terms containing x so that you get the ratio x/d instead of just x and d separately. For example $x + d = x(1 +d/x)$ Cancel any terms that appear in the numerator and denominator. Then set $d/x=0$ and see what is left.

On edit: If that doesn't work, you will have to use a Taylor expansion for small d/x and keep the leading term.

Last edited: Feb 28, 2017
5. Feb 28, 2017

### Staff: Mentor

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