Give the advantages and disadvantages of join family and nuclear family ...( give any 5 points of advantages and disadvantages for the two topic..
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Answers
Answer:
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
\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}}.
A remarkable fact about this equation is that the flux is independent of the size of the spherical surface. This can be directly attributed to the fact that the electric field of a point charge decreases as 1\text{/}{r}^{2} with distance, which just cancels the {r}^{2} rate of increase of the surface area.
Electric Field Lines Picture
An alternative way to see why the flux through a closed spherical surface is independent of the radius of the surface is to look at the electric field lines. Note that every field line from q that pierces the surface at radius {R}_{1} also pierces ward from the center in all eight directions.
Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction is equal. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction from outside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric flux through the surfaces.
You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typical field line enters the surface at d{A}_{1} and leaves at d{A}_{2}. Every line that enters the surface must also leave that surface. Hence the net “flow” of the field lines into or out of the surface is zero ((Figure)(a)). The same thing happens if charges of equal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). A surface that includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surface encloses the same amount of charge (part (c)).
Understanding the flux in terms of field lines. (a) The electric flux through a closed surface due to a charge outside that surface is zero. (b) Charges are enclosed, but because the net charge included is zero, the net flux through the closed surface is also zero. (c) The shape and size of the surfaces that enclose a charge does not matter because all surfaces enclosing the same charge have the same flux.
Figure a shows an irregular 3 dimensional shape labeled S. A small circle with a plus sign, labeled q is outside it. Three arrows labeled vector E originate from q and pass through S. The patches where the arrows pierce the surface of S are highlighted. The
Statement of Gauss’s Law