The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\times 10^{-5}~\text{N}\cdot\text{m}^2/\text{C}3×10 ​−5 ​​ N⋅m ​2 ​​ /C when the area is parallel to the sheet of charge. Find the charge density on the sheet.

Respuesta :

Answer:

Explanation:

Given

side of square shape [tex]a=5\ cm[/tex]

Electric flux [tex]\phi =3\times 10^{-5}\ N.m^2/C[/tex]

Permittivity of free space [tex]\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}[/tex]

Flux is given by

[tex]\phi =EA\cos \theta [/tex]

where E=electric field strength

A=area

[tex]\theta [/tex]=Angle between Electric field and area vector

[tex]E=\frac{\phi }{A\cos (0)}[/tex]

[tex]E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}[/tex]

[tex]E=0.012\ N/C[/tex]

and Electric field  by a uniformly charged sheet is given by

[tex]E=\frac{\sigma }{2\epsilon_0}[/tex]

where [tex]\sigma[/tex]=charge density

[tex]=\frac{\sigma }{\epsilon_0}[/tex]

[tex]\sigma =0.012\times 8.85\times 10^{-12}[/tex]

[tex]\sigma =2.12\times 10^{-13}\ C/m^2[/tex]