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

`E=576.5V/m`

Step-by-step 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=(\rho)/(e_0)`

`E=\frac{5.10*10^(-9)}{8.85x10{-12}}`

`E=576.5V/m`