A closed Gaussian surface encloses a charge ‘q’ if the charge moves within the Gaussian surface, the physical quantity which varies is -
Electric Field
Electric Flux
Charge
Area of Gaussian Surface
Answers
Explanation:
We can now determine the electric flux through an arbitrary closed surface due to an arbitrary charge distribution. We found that if a closed surface does not have any charge inside where an electric field line can terminate, then any electric field line entering the surface at one point must necessarily exit at some other point of the surface. Therefore, if a closed surface does not have any charges inside the enclosed volume, then the electric flux through the surface is zero. Now, what happens to the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to this question.
To get a feel for what to expect, let’s calculate the electric flux through a spherical surface around a positive point charge q, since we already know the electric field in such a situation. Recall that when we place the point charge at the origin of a coordinate system, the electric field at a point P that is at a distance r from the charge at the origin is given by
{\stackrel{\to }{\text{E}}}_{P}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{r}^{2}}\stackrel{^}{\text{r}},
where \stackrel{^}{\text{r}} is the radial vector from the charge at the origin to the point P. We can use this electric field to find the flux through the spherical surface of radius r, as shown in (Figure).
A closed spherical surface surrounding a point charge q.
A sphere labeled S with radius R is shown. At its center, is a small circle with a plus sign, labeled q. A small patch on the sphere is labeled dA. Two arrows point outward from here, perpendicular to the surface of the sphere. The smaller arrow is labeled n hat equal to r hat. The longer arrow is labeled vector E.
Then we apply \text{Φ}={\int }_{S}\stackrel{\to }{E}·\stackrel{^}{n}\phantom{\rule{0.2em}{0ex}}dA to this system and substitute known values. On the sphere, \stackrel{^}{n}=\stackrel{^}{r} and r=R, so for an infinitesimal area dA,
d\text{Φ}=\stackrel{\to }{E}·\stackrel{^}{n}\phantom{\rule{0.2em}{0ex}}dA=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{q}{{R}^{2}}\stackrel{^}{r}·\stackrel{^}{r}\phantom{\rule{0.2em}{0ex}}dA=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{q}{{R}^{2}}\phantom{\rule{0.2em}{0ex}}dA.
We now find the net flux by integrating this flux over the surface of the sphere:
\text{Φ}=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{q}{{R}^{2}}{\oint }_{S}\phantom{\rule{0.2em}{0ex}}dA=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{q}{{R}^{2}}\left(4\pi {R}^{2}\right)=\frac{q}{{\epsilon }_{0}}.
where the total surface area of the spherical surface is 4\pi {R}^{2}. This gives the flux through the closed spherical surface at radius r as
\text{Φ}=\frac{q}{{\epsilon }_{0}}.
Answer:
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Explanation:
When dipole moment vector is parallel to electric field vector.