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This is an inclined surface having and unit vector A. The perpendicular and horizontal surfaces are the components of the inclined surface. We want to find out that how much flux will pass through the inclined surface.

My question is that why don't we take θ at the point I have marked red? Why have we taken θ on the other point?

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You are perfectly free to call the other angle $\theta$, so long as you are consistent. Then the area of interest would be written $A \sin(\theta)$, and the other $\theta$ marked in the diagram would have to be relabeled $\pi/2-\theta$.

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  • $\begingroup$ But that is the problem. If i consider the other angle (red) to be θ then I will have to write the Electric flux as: Electric flux = EAsinθ And this mathematical form is wrong for Electric flux. $\endgroup$ – Rafique Apr 26 '13 at 0:00
  • $\begingroup$ @MuhammadRafique Why is it wrong? The electric flux is whatever it happens to be. Where does it state that only cosines can appear in the formula? $\endgroup$ – user10851 Apr 26 '13 at 0:34
  • $\begingroup$ Electric flux = EAsinθ and Electric flux = EAcosθ will give two seperate answers, wouldn't it? $\endgroup$ – Rafique Apr 26 '13 at 0:43
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    $\begingroup$ @Muhammad Rafique: Not if the $\theta$s are different. You're thinking of $\cos\theta$ in the dot product, but that $\theta$ is the angle between the two dotted vectors. If we take your red angle, let's call it $\theta'$, then the angle between A and E is $\pi/2-\theta'$, and $\cos(\pi/2-\theta') = \sin\theta'$. It's all consistent. $\endgroup$ – Stan Liou Apr 26 '13 at 0:45
  • $\begingroup$ oh yeahh.....:) $\endgroup$ – Rafique Apr 26 '13 at 1:13
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Imagine taking the inclined surface to be close to horizontal. In this case, the angle between $\mathbf A$ and the horizontal would be approximately $90^\circ$, while the angle you have indicated in red would be approximately $0^\circ$, so they can't possibly be the same.

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  • $\begingroup$ I don't understand. what do you mean by "take inclined surface to be close to horizontal"? $\endgroup$ – Rafique Apr 26 '13 at 0:02
  • $\begingroup$ In the original picture, I think the intent was for the angle $\theta$ to represent the angle between the vector $\mathbf A$, and the ground on which the inclined plane rests. In this case, $\theta=90^\circ$ then the surface of the incline is horizontal (parallel to the ground), and $\theta = 0^\circ$ when the surface of the incline is vertical. The (red) angle between the inclined plane and the horizontal has the opposite behavior; it's $0^\circ$ when the inclined plane is horizontal, and $90^\circ$ when the inclined plane is vertical. So it can't be the same as $\theta$. $\endgroup$ – joshphysics Apr 26 '13 at 1:47

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