Answer:
18.1 V
Explanation:
The electric field between two parallel plates is given by the equation:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the charge surface density
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
For the plates in this problem,
[tex]\sigma = 2.0 nC/m^2 = 2.0\cdot 10^{-9} C/m^2[/tex]
So, the magnitude of the electric field is
[tex]E=\frac{2.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=226.0 V/m[/tex]
Now we can find the potential difference between the plates, which is given by
[tex]\Delta V = E d[/tex]
where
d = 8.0 cm = 0.08 m is the separation between the plates
Substituting,
[tex]\Delta V=(226.0)(0.08)=18.1 V[/tex]
Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces.
The potential difference between the plates is 18.072 Volts.
Given that the distance d between the plates is 8.0 cm, the magnitude of the surface charge density is 2.0 nC/m2.
The magnitude of the electric field between the plates is calculated as given below.
[tex]E = \dfrac {\sigma }{\varepsilon_0}[/tex]
Where E is the electric field, [tex]\varepsilon_0[/tex] is the permittivity of vacuum, and [tex]\sigma[/tex] is the surface charge density.
[tex]E = \dfrac {2.0 \times 10^{-9}}{8.85\times 10^{-12}}[/tex]
[tex]E = 225.9 \;\rm V/m[/tex]
The potential difference between the plates that are at a distance d apart is given below.
[tex]\Delta V = Ed[/tex]
[tex]\Delta V = 225.9 \times 0.08[/tex]
[tex]\Delta V = 18.072\;\rm V[/tex]
Hence we can conclude that the potential difference between the plates is 18.072 Volts.
To know more about the potential difference, follow the link given below.
https://brainly.com/question/1313684.
