Respuesta :
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² = [tex]47\times 10^{-9}\ C/m^2[/tex]
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
[tex]E=\dfrac{\sigma}{2\epsilon_0}[/tex]
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
[tex]E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}[/tex]
Now, plug in [tex]47\times 10^{-9}\ C/m^2[/tex] for 'σ' and [tex]8.85\times 10^{-12}\ F/m[/tex] for [tex]\epsilon_0[/tex] and solve for the electric field. This gives,
[tex]E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C[/tex]
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
