Question

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surface charge density for each plate has magnitude 47.0 nC/m2, what is the magnitude of E⃗ in the region between the plates?

Answer 1

5308.34 N/C

Step-by-step explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² =
`47* 10^(-9)\ C/m^2`

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:

`E=(\sigma)/(2\epsilon_0)`

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,

`E_(between)=E+E=2E=(2\sigma)/(2\epsilon_0)=(\sigma)/(\epsilon_0)`

Now, plug in
`47* 10^(-9)\ C/m^2` for 'σ' and
`8.85* 10^(-12)\ F/m` for
`\epsilon_0` and solve for the electric field. This gives,

`E_(between)=(47* 10^(-9)\ C/m^2)/(8.854* 10^(-12)\ F/m)\\\\E_(between)= 5308.34\ N/C`

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C