Net electric field question almost have it but can't get one part of it

[itsagulati]

Homework Statement

Two charges, -26 µC and +5 µC, are fixed in place and separated by 1.3 m.
(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does not necessarily lie between the two charges.)

Homework Equations

E = k*q/r^2

The Attempt at a Solution

26e-6k / d^2 = 5e-6k / (1.3 - d)^2

i solved for that and got d = .9037 and d = 2.32 ...since i need them pointing in the same direction its the one that is not between them. i understand that...and it all seems right.

Does anyone get something different?

It is an odd #'d problem so i looked at the question in the book and was able to work it to get the right answers for the book version, but my online #s are different and while i work it very similarly and it all makes logical sense. I get them wrong :(

Any tips?

 

[learningphysics]

The question asks for the distance from the positive charge... so this is not 'd'... from your formula d is the distance from the negative charge...

Although your equation is technically correct... I'd advise first finding the region where the field is 0, then choosing your variable appropriately... There was a recent thread on this and Doc Al gave a great explanation.

 

[itsagulati]

I saw that and read it and hten posted this one. Based on what i picked up in that thread, I ended up with the above.

However, based on what you said I assume if we drew it out on a line...with the negative charge as our starting point...2.32m out would be what I said was the answer. However, since there is a distance from the negative charge to the positive charge of 1.3m...then 2.32m- 1.3m = 1.02m? Does this sound right?

Thanks!

 

[learningphysics]

I saw that and read it and hten posted this one. Based on what i picked up in that thread, I ended up with the above.

However, based on what you said I assume if we drew it out on a line...with the negative charge as our starting point...2.32m out would be what I said was the answer. However, since there is a distance from the negative charge to the positive charge of 1.3m...then 2.32m- 1.3m = 1.02m? Does this right?

Thanks!

Yes, 1.02m is right. That's cool... I thought since you used (1.3 - d)^2 instead of (d - 1.3)^2, you were expecting d to be less than 1.3... both equations are the same...

As long as you know what's happening, it's all good.