Question

If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the square. For the uniform electric field normal to the surface, the flux through the surface is____________the area of this surface. Therefore, Φsquare is ________ ϕcircle .

Answer 1

Answer:

The area of this circle is $(\frac{\pi}{2} )$  the area of the square.

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Therefore, Φsquare is $(\frac{2}{\pi} )$ ϕcircle

Step-by-step explanation:

Area of the circle is given by;

$A_c = \frac{\pi d^2}{4}$

Area of the square is given by;

$A_s = L^2$

relationship between the edge length of the square, d, and length of its side, L,

$d = \sqrt{L^2 + L^2} \\\\d = \sqrt{2L^2}$

But area of the square , $A_s = L^2$

$d = \sqrt{2A_s}$

Then, the area of the square in terms of the edge length is given by;

$A_s = \frac{d^2}{2}$

Area of the circle in terms of area of the square is given by;

$A_c = \frac{\pi d^2}{4} = \frac{\pi}{2}(\frac{d^2}{2} )\\\\But \ A_s = \frac{d^2}{2} \\\\A_c = \frac{\pi}{2}(\frac{d^2}{2} )\\\\A_c = \frac{\pi}{2}(A_s )$

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Ф = E.A

Flux through the surface of the circle is given by;

$\phi _{circle} = E.(\frac{\pi d^2}{4})$

Flux through the surface of the square is given by;

$\phi _{square} = E.(\frac{d^2}{2} )\\\\\phi _{square} =E.(\frac{d^2}{2} ).(\frac{\pi}{2} ).(\frac{2}{\pi} )\\\\\phi _{square} =E.(\frac{\pi d^2}{4} ).(\frac{2}{\pi} )\\\\\phi _{square} =(\phi _{circle}).(\frac{2}{\pi} )$

Therefore, Φsquare is $(\frac{2}{\pi} )$ ϕcircle