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PFStudent
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Homework Statement
4. In Fig. 23-28, a butterfly net is in a uniform electric field of magnitude [itex]{E = 3.0}[/itex] mN/C. The rim, a circle of radius [itex]a[/itex] = 11 cm, is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.
http://img220.imageshack.us/img220/4193/hrw72328xx2.gif
[Fig 23-28]
Homework Equations
Gauss' Law
[tex]
{{\Phi}_{E}} = {{\oint}{\vec{E}}{\cdot}{d{\vec{A}}}}
[/tex]
The Attempt at a Solution
At first I thought this was a simple problem. I reasoned that the net electric flux ([itex]{{\Phi}_{E}}[/itex]) was zero because all the electric field lines that enter leave also.
However, the book answer was not zero and so proved that answer wrong.
Then I thought to just evaluate the electric flux noting that the the [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] point in the same direction so that,
[tex]
{{\Phi}_{E}} = {{\oint}{\vec{E}}{\cdot}{d{\vec{A}}}}
[/tex]
[tex]
{{\Phi}_{E}} = {{\oint}{|\vec{E}|}{|d{\vec{A}}}|}{{cos}{\theta}}
[/tex]
Where,
[tex]
{\theta} = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}degrees
[/tex]
Since, [itex]{\vec{E}}[/itex] and [itex]{\vec{A}}[/itex] point in the same direction.
So,
[tex]
{{\Phi}_{E}} = {|\vec{E}|}{{\oint}{|d{\vec{A}}}|}
[/tex]
[tex]
{{\Phi}_{E}} = {|\vec{E}|}{|{\vec{A}}}|}
[/tex]
Where [itex]A > 0[/itex], so letting, [itex]{|{\vec{A}}}|} = A[/itex]. Noting that [itex]A = {{\pi}{{a}^{2}}}[/itex].
[tex]
{{\Phi}_{E}} = {|\vec{E}|}{A}
[/tex]
[tex]
{{\Phi}_{E}} = {|\vec{E}|}{{\pi}{{a}^{2}}}
[/tex]
However the book answer is the negative of the above.
So my question is, why the negative answer?
Thanks,
-PFStudent
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