Question

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?

Answer 1

Answer:

$E=576.5V/m$

Explanation:

From the question we are told that:

Length $l=56.0cm=0.56m$

Distance apart $d=7.0mm=0.007m$

Electron Transferred $n=10^{-10}$

Therefore

Total Charge

Since Charge on each electron is

$e=1.602*10^{-19}$

Therefore

$T=1.602*10^{-19} *10^{10}$

$T=1.602*10^{-9}$

Generally the equation for Charge density is mathematically given by

$\rho=T/A$

Where

Area

$A=0.56*0.56$

$A=0.3136$

Therefore

$\rho=1.602*10^{-9}/0.3136$

$\rho=5.10*10^{-9}$

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

$E=\frac{\rho}{e_0}$

$E=\frac{5.10*10^{-9}}{8.85x10{-12}}$

$E=576.5V/m$