# Dot product vs trigonometry in Gauss' law

I'm currently writing my EP on various physical equations including Maxwell's equations, and I had to justify using the dot product of the normal unit vector and the electric field in the integral version. However, I can't think of a reason for not using trigonometry as opposed to the |a||b|cos<(a,b). Any clarification or explanation is very welcome.

HallsofIvy
Homework Helper
I am not sure I understand your question. What, exactly, do you mean by "use trigonometry"? It seems to me that if you were to "use trigonometry" (in the usual sense) you would get exactly the same result as using the dot product.

What's the trigonometry you meant in Guass' Law?

Sorry, I just realised that they would give the same result :P
But why is the dot product written instead of trig? Is it because it's easier to write out in an equation?

ZapperZ
Staff Emeritus
Sorry, I just realised that they would give the same result :P
But why is the dot product written instead of trig? Is it because it's easier to write out in an equation?

This is getting to be rather silly.

Don't be lazy. Write down the exact equations that you are talking about, because it is obvious that the rest of us have no idea what you are talking about. This forum has the ability to use LaTex math formatting. Use that and show us exactly the type of equations you are referring to.

Otherwise, we have this rather puzzling description from you which makes very little sense!

Zz.

I'm new to the forum so I apologize. However, there is no need to be quite that rude to me.
I am not being lazy, I'm being ignorant :)
Have a nice day now

Nugatory
Mentor
Sorry, I just realised that they would give the same result :P
But why is the dot product written instead of trig? Is it because it's easier to write out in an equation?
I presume you're asking why we write the integrand as the dot product of the E vector and the normal unit vector, instead of using the expression you posted (product of their magnitudes and the angle between them)?

It's because the dot product of vectors can be calculated without knowing the angle between them, and in more advanced problems it can be difficult or impossible to find this angle. This is one of many nice mathematical properties that make the dot product more generally useful than a one-off trig-based calculation.

You probably won't see how much more powerful the dot product is until you get into linear algebra and non-trivial coordinate transforms. Until then, you may have to take our word for it that's it a better tool and that you'll want to get comfortable with it.

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