An Electrostatics Problem - Eletric Field

[Yukimi]

Homework Statement

Question 2.2
http://www.studyjapan.go.jp/en/toj/pdf/08-007.pdf
The picture is in the next page of the problem.

• k is the Coulomb constant.
• q+ is the charge at point A.
• q- is the charge at point B.
• AC = BC = CD = a

Homework Equations

E = kq/d²
a² = b² + c²

The Attempt at a Solution

I put the vectors on the force line at AD and DB, adding the vectors we get the eletric field at point D, the problem is that when I use a² = b² + c² (observe that is a right triangle) I get the result Kq/a, answer D, but the correct answer is B. Where is wrong?

I have just learned Electrostatics, maybe it is a simple problem, but I can not solve.

Thank you.

 

[tiny-tim]

Welcome to PF!

Hi Yukimi! Welcome to PF!

I put the vectors on the force line at AD and DB, adding the vectors we get the eletric field at point D, the problem is that when I use a² = b² + c² (observe that is a right triangle) I get the result Kq/a, answer D, but the correct answer is B. Where is wrong?

Show us how you got Kq/a.

 

[Yukimi]

As I said, I got by doing the addition of the vectors from A to D (force line from charge q+) and from D to B (force line from charge q-), I did that because the final vector will be the electric field at point D (Ed), using the pythagorean theorem (it is a right triangle, because DC and AB are perpendicular) I got AD = AB = a times square root of 2, using the pythagorean theorem one more time I got Ed = Kq/a.

 

[tiny-tim]

I don't understand how you avoided getting a square on the bottom.

 

[Yukimi]

Sorry, I don't understand, are you saying about the equation using the pythagorean theorem? If that is the case, here is how I did:

AB² + CD² = AD²
a² + a² = AD²
a*root 2 = AD

Ead² + Edb² = Ed²
(kq/a*root 2)² + (kq/a*root 2)² = Ed²
2[(kq)²/2a²] = Ed²
kq/a = Ed

 

[gneill]

Ah. The magnitude of the field strength for a charge at distance a√2 is

E = k*q/(a√2)² = k*q/(2a²)

That's the magnitude of the field due to one charge. Resolve this into components via geometry for both of the charge sources. Add the like components to find the net field.

 

[tiny-tim]

Ead² + Edb² = Ed²

(kq/a*root 2)² + (kq/a*root 2)² = Ed²

(kq/(a*root 2)²)² + (kq/(a*root 2)²)² = Ed²

 

[Yukimi]

wow What stupid mistake!

Thank you both! ^_____^