Quick clarification/question on Electric Field at a point

[Pablo1122]

Homework Statement

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?

Homework Equations

v = kq/r
E=kq/r^2
F=qe

The Attempt at a Solution

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

Alright, so I've done the part a) in the following manner.[/B]

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

 

[mjc123]

Your pictures are very hard to read. Please take the trouble to type out your working. It is disrespectful to forum members to expect them to make the effort to read your scrawl if you can't make the effort to type it legibly. Please read the guidelines: https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/
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.

 

[Pablo1122]

Your pictures are very hard to read. Please take the trouble to type out your working. It is disrespectful to forum members to expect them to make the effort to read your scrawl if you can't make the effort to type it legibly. Please read the guidelines: https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/
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.

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.

 

[kuruman]

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.

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.

 

[gneill]

I'm sorry if I was disrespectful I just don't know how to go about writing it.

You can type equations using normal text making using of the x₂ and x² buttons in the edit panel top bar to produce subscripts and superscripts, the special characters including the Greek alphabet and many handy math symbols via a menu accesses via the $\Sigma$ button also in the top bar, or by using LaTeX syntax. See: LaTeX