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	<title>
	Comments on: Vector Calculus: Understanding Flux	</title>
	<atom:link href="https://betterexplained.com/articles/flux/feed/" rel="self" type="application/rss+xml" />
	<link>https://betterexplained.com/articles/flux/</link>
	<description>Math lessons that click</description>
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		<title>
		By: Madison Adriana		</title>
		<link>https://betterexplained.com/articles/flux/#comment-567719</link>

		<dc:creator><![CDATA[Madison Adriana]]></dc:creator>
		<pubDate>Sat, 30 Nov 2019 02:19:33 +0000</pubDate>
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					<description><![CDATA[&lt;p&gt;this was so helpful bless your soul. thank you&lt;/p&gt;]]></description>
			<content:encoded><![CDATA[<p>this was so helpful bless your soul. thank you</p>
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		<title>
		By: Soumitra		</title>
		<link>https://betterexplained.com/articles/flux/#comment-567631</link>

		<dc:creator><![CDATA[Soumitra]]></dc:creator>
		<pubDate>Fri, 20 Sep 2019 20:50:35 +0000</pubDate>
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					<description><![CDATA[&lt;p&gt;Thanks.&lt;br&gt;I came from divergence&lt;/p&gt;]]></description>
			<content:encoded><![CDATA[<p>Thanks.<br />I came from divergence</p>
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		<title>
		By: kalid		</title>
		<link>https://betterexplained.com/articles/flux/#comment-567376</link>

		<dc:creator><![CDATA[kalid]]></dc:creator>
		<pubDate>Fri, 22 Feb 2019 21:43:58 +0000</pubDate>
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					<description><![CDATA[In reply to &lt;a href=&quot;https://betterexplained.com/articles/flux/#comment-567375&quot;&gt;Victor Iglesias Castan&lt;/a&gt;.

&lt;p&gt;Good question. The vector from the center of the sphere (0, 0, 0) to your current position (x, y, z) is always perpendicular [normal] to the surface of the sphere. (Assume a sphere centered at the origin.)&lt;/p&gt;&lt;p&gt;If you&#039;re on the surface of a sphere at position (1, 0, 0), the normal vector points along (1, 0, 0).&lt;/p&gt;&lt;p&gt;If you&#039;re on a larger sphere at position (1, 1, 1), your normal vector points along (1, 1, 1).&lt;/p&gt;]]></description>
			<content:encoded><![CDATA[<p>In reply to <a href="https://betterexplained.com/articles/flux/#comment-567375">Victor Iglesias Castan</a>.</p>
<p>Good question. The vector from the center of the sphere (0, 0, 0) to your current position (x, y, z) is always perpendicular [normal] to the surface of the sphere. (Assume a sphere centered at the origin.)</p>
<p>If you&#8217;re on the surface of a sphere at position (1, 0, 0), the normal vector points along (1, 0, 0).</p>
<p>If you&#8217;re on a larger sphere at position (1, 1, 1), your normal vector points along (1, 1, 1).</p>
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		<item>
		<title>
		By: Victor Iglesias Castan		</title>
		<link>https://betterexplained.com/articles/flux/#comment-567375</link>

		<dc:creator><![CDATA[Victor Iglesias Castan]]></dc:creator>
		<pubDate>Fri, 22 Feb 2019 18:31:30 +0000</pubDate>
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					<description><![CDATA[&lt;p&gt;Great article, Kalid, congratulations. I have one question about this article, when you say:&lt;/p&gt;&lt;p&gt;&lt;/p&gt;&lt;blockquote&gt;the normal vector is in the same direction as r,&lt;/blockquote&gt;&lt;p&gt;&lt;/p&gt;&lt;p&gt;What do you mean with &lt;i&gt;r&lt;/i&gt;? The radius of the sphere?&lt;/p&gt;&lt;p&gt;On the other hand, I suppose that it has been determined experimentally that the electrostatic flux in a given surface matches with the dot product between the vector &quot;field&quot; and the vector &quot;surface&quot; which direction is perpendicular to that surface and its norm is the area surface. For me, that &quot;coincidence&quot; is fascinating.&lt;/p&gt;&lt;p&gt;(Excuse me for my English, I&#039;m not a native speaker)&lt;/p&gt;]]></description>
			<content:encoded><![CDATA[<p>Great article, Kalid, congratulations. I have one question about this article, when you say:</p>

<blockquote><p>the normal vector is in the same direction as r,</p></blockquote>
<p>What do you mean with <i>r</i>? The radius of the sphere?</p>
<p>On the other hand, I suppose that it has been determined experimentally that the electrostatic flux in a given surface matches with the dot product between the vector &#8220;field&#8221; and the vector &#8220;surface&#8221; which direction is perpendicular to that surface and its norm is the area surface. For me, that &#8220;coincidence&#8221; is fascinating.</p>
<p>(Excuse me for my English, I&#8217;m not a native speaker)</p>
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