A butterfly net is in a uniform electric field of magnitude E = 3.0 mN/C. The rim, a circle of radius a = 11 cm, is aligned perpendicular to the field. The net contains no net charge. Find the electric flux through the netting.
Solution:
We can use the electric flux equation to figure out the flux through the netting.
\Phi= \displaystyle\oint \vec E \cdot (d\vec A)-\vec i
\Phi= \displaystyle\oint (0.003) \cdot (d\vec A)-\vec i
\Phi= -0.003\displaystyle\oint dA
\Phi= -0.003A
\Phi= -0.003(\pi)(r^2)
\Phi=-0.003(\pi)(0.11^2)
\Phi=-1.14\times 10^{-4}\dfrac{N\cdot m^2}{C}
\Phi= \displaystyle\oint \vec E \cdot (d\vec A)-\vec i
\Phi= \displaystyle\oint (0.003) \cdot (d\vec A)-\vec i
\Phi= -0.003\displaystyle\oint dA
\Phi= -0.003A
\Phi= -0.003(\pi)(r^2)
\Phi=-0.003(\pi)(0.11^2)
\Phi=-1.14\times 10^{-4}\dfrac{N\cdot m^2}{C}
Our answer is negative because the question asks us to find the electric flux “through the netting” meaning the direction in which things can enter the net. Thus, it’s opposite to the direction of the electric field.