- #1

Bashyboy

- 1,421

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## 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:

[itex]\vec{E_A}=k_e\frac{5q}{a^2}\widehat{i}[/itex]

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

[itex]\vec{E_C}=k_e\frac{7q}{a^2}\widehat{j}[/itex]

Summing the effects of the each electric field together:

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

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

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