Question

The square and circle in FIGURE Q24.3 are in the same uniform field. The diameter of the circle equals the edge length of the square. Is ${\Phi }_{square}$larger than, smaller than, or equal to ${\Phi }_{circle}$? Explain.

Short answer

The electric flux through the square is larger than the electric flux through the square.

Step-by-step solution

Given information and formula used  

Given :

The square and circle are in the same uniform field.

The diameter of the circle = the edge length of the square.

Theory used :

The amount of electric field that travels through a closed surface is referred to as the electric flux.

The electric field through a surface is related to the charge inside the surface, according to Gauss's law. When the electric field is homogeneous, we compute the electric flow using equation :

${\varphi }_e=\overrightarrow{E}·\overrightarrow{A}$

Determining if Φsquare larger than, smaller than, or equal to Φcircle

The diameter of a circle is $L$, which is the same as the length of the square. The circle's area is:

${\overrightarrow{A}}_{circle}=\mathrm{\pi }{\left(\frac{\mathrm{L}}{2}\right)}^2=0.785L^2$ ; so

${\varphi }_{circle}=E(0.785L^2)=\underline{0.785E{L}^2}$

is the electric field through the circle.

Now, the square's area is :

$A_{square}=L\times L=L^2.$

And the electric field through the square is :

role="math" localid="1649315945211" ${\varphi }_{square}=E\left(L^2\right)=\underline{E{L}^2}$

The electric flux through the square is larger than the electric flux through the square, as evidenced by the data.