Finding The Effect of Several Electrics Fields

[Bashyboy]

Homework Statement

Four charged particles are at the corners of a square of side a as shown in the figure below. (Let A = 5, B = 2, and C = 7.)

Homework Equations

The Attempt at a Solution

Well, I first found the electric due to each particle individually:

$\vec{E_A}=k_e\frac{5q}{a^2}\widehat{i}$

$\vec{E_B}=k_e \large[ \frac{2q~cos(45°)}{a^2}\widehat{i}+\frac{2q~sin(45°)}{a^2}\widehat{j}]$

$\vec{E_C}=k_e\frac{7q}{a^2}\widehat{j}$

Summing the effects of the each electric field together:

$\vec{E_{tot}}=k_e \large[(\frac{5q+2q\cos{45°}}{a^2}\widehat{i}+(\frac{2q \sin{45°}+7q}{a^2}\widehat{j}$

After simplifying, I found the magnitude of the electric field at point q, that the three particles create, to be $10.58 \cdot \frac{q}{a^2}$; however, the true answer is, $9.59 \cdot \frac{q}{a^2}$ What did I do wrong?

 

[Doc Al]

You did not include a diagram or describe what you are asked to find.

 

[Bashyboy]

Sorry. I just attached one.

 

[Doc Al]

OK, what's the distance between B and q?

 

[Bashyboy]

Wouldn't it be $\sqrt{2}a$?

 

[Bashyboy]

I figured it would be better to resolve the electric field of B into its components.

 

[Doc Al]

Wouldn't it be $\sqrt{2}a$?

Right.

I figured it would be better to resolve the electric field of B into its components.

Nothing wrong with that, but you must use the correct distance to calculate the field.

 

[Bashyboy]

Well, to get from point B to point q, don't I have to go a units to right and a units north? What are the correct distances?

 

[Doc Al]

Well, to get from point B to point q, don't I have to go a units to right and a units north? What are the correct distances?

You just gave the correct distance in your earlier post. Use it!

 

[Bashyboy]

Oh I see, I am mixing the idea of resolving charges and distances together.