) each plate of a parallel-plate air-filled capacitor has an area of 0.0020 , and the separation of the plates is an electric field of is present between the plates. what is the surface charge density on the plates? (ε0 = 8.85 × 10-12 c2/n • m2)

I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
[tex]\sigma=\varepsilon_0 \frac{V}{d}[/tex]
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
[tex]V=Ed[/tex]
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
[tex]\sigma=\varepsilon_0 \frac{Ed}{d}=\varepsilon_0E\\ \sigma= 8.85\cdot10^{-12}\cdot 2.1\cdot 10^6= 0.0000185\frac{c}{m^2}[/tex]
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them interchangeably.

keshavgandhi04 keshavgandhi04 · Answer 2 · 2022-02-23T11:07:19+01:00

The surface charge density on plates when the electric field of 2.1 * 10^6 N/C is present is [tex]18.585*10^{-6}[/tex][tex]c/m^{2}[/tex]

What is Surface Charge Density?

Surface charge density is the charge per unit area which can be measured in coulombs per square meter.

Now, we know that the surface charge density for parallel plate capacitor is given by:

[tex]\sigma=\epsilon_{0}\cdot(\frac{v}{d})[/tex] ...........(i)

Where, v= Voltage and d= seperation between the plates

As we know that [tex]v=E\cdot{d}[/tex]............(ii)

So, by considering the equations (i) and (ii) we get,

[tex]\sigma=\epsilon_{0}\cdot{E}[/tex]

Putting the value in the equations we get,

[tex]\sigma[/tex]=[tex](8.85*10^{-12})[/tex][tex](2.1*10^{6} )[/tex]

=[tex]18.585*10^{-6}[/tex] [tex]c/m^{2}[/tex]

To know more about Surface Charge Density

https://brainly.com/question/8966223