Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons are moved from one plate to the other, what is the electric field between the plates?

Respuesta :

Answer:

[tex]E=576.5V/m[/tex]

Explanation:

From the question we are told that:

Length [tex]l=56.0cm=0.56m[/tex]

Distance apart [tex]d=7.0mm=0.007m[/tex]

Electron Transferred [tex]n=10^{-10}[/tex]

Therefore

Total Charge

Since Charge on each electron is

[tex]e=1.602*10^{-19}[/tex]

Therefore

[tex]T=1.602*10^{-19} *10^{10}[/tex]

[tex]T=1.602*10^{-9}[/tex]

Generally the equation for Charge density is mathematically given by

[tex]\rho=T/A[/tex]

Where

Area

[tex]A=0.56*0.56[/tex]

[tex]A=0.3136[/tex]

Therefore

[tex]\rho=1.602*10^{-9}/0.3136[/tex]

[tex]\rho=5.10*10^{-9}[/tex]

Generally the equation for Electric Field in the capacitor is mathematically given by

[tex]E=\frac{\rho}{e_0}[/tex]

[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]

[tex]E=576.5V/m[/tex]