(a) 5310.7 V/m
The magnitude of the electric field between two parallel plates is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
In this problem,
[tex]\sigma = 47.0 nC/m^2 = 47.0 \cdot 10^{-9} C/m^2[/tex]
So the electric field here is
[tex]E=\frac{47.0\cdot 10^{-9} C/m^2}{8.85\cdot 10^{-12}F/m}=5310.7 V/m[/tex]
(b) 116.8 V
The potential difference between the two plates is given by
[tex]V= Ed[/tex]
where
E is the magnitude of the electric field
d is the separation between the plates
Here we have
E = 5310.7 V/m
d = 2.20 cm = 0.022 m
So the potential difference is
[tex]V=(5310.7 V/m)(0.022m)=116.8 V[/tex]
(c) The electric field does not change
Explanation:
As stated in part (a), the magnitude of the electric field is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
as we can see, the value of E depends only on the surface charge density, which is kept constant in this case, so the value of the electric field strength does not change.
(d) The potential difference doubles (233.6 V)
In this situation, the separation between the plates is doubled, so:
d' = 2 d
The potential difference depends linearly on the separation between the plates:
V = Ed
where
E is the magnitude of the electric field (which is kept constant)
d is the separation between the plates
So the new potential difference will be
[tex]V' = E(2d) = 2 (Ed) = 2 V[/tex]
which means that the potential difference will double:
[tex]V'=2 (116.8 V)=233.6 V[/tex]
