Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 )if the surface charge density for each plate has magnitude 47.0 nc/m2, what is the magnitude of e_field in the region between the plates? (b)what is the potential difference between the two plates? (c)if the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field? (d)if the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the potential difference?

(a) 5310.7 V/m

The magnitude of the electric field between two parallel plates is given by

where

is the surface charge density

is the vacuum permittivity

In this problem,

So the electric field here is

(b) 116.8 V

The potential difference between the two plates is given by

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

(c) The electric field does not change

Explanation:

As stated in part (a), the magnitude of the electric field is given by

where

is the surface charge density

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

which means that the potential difference will double:

true is the correct answer

hope i