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Quick Answer: 

$\dfrac{kQ}{l^2}, 30^\circ \textrm{ below the positive } x \textrm{ axis.}$

Giancoli 7th Edition, Chapter 16, Problem 33

(6:14)

Chapter 16, Problem 33 is solved.

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Comments

Hi,

Could you explain why you used 30 degree angles instead of 60?

Thank you.

Hi ctlawson, as part of our job in this question we need to break the electric fields into 'x' and 'y' components. It would be perfectly OK to use $60^\circ$. The difference is that you would have to think about which trig. function is the right one to use. For example, when I calculate $E_{Bx}$ using $\cos(30^\circ)$ you could instead use $\sin(60^\circ)$. It's just a matter of personal preference.

Cheers,
Mr. Dychko

Thank you, Mr. Dychko.