88.2k views
2 votes
A square plate of copper with 47.0 cm sides has no net charge and is placed ina region of uniform electric field of 75.0 kN/C directed perpendicularly to the plate.

(a) Find the charge density of each face of the plate.
nC/m2
nC/m2

(b) Find the total charge on each face.
nC
nC

User MkRabbani
by
6.4k points

1 Answer

3 votes

Answer:

(a) Charge density σ=6.6375×10²nC/m²

(b) Total charge Q=1.47×10²nC

Step-by-step explanation:

Given Data

A=47.0 cm =0.47 m

Electric field E=75.0 kN/C

To find

(a) Charge density σ

(b)Total Charge Q

Solution

For (a) charge density σ

From Gauss Law we know that

Φ=Q/ε₀.......eq(i)

Where

Φ is electric flux

Q is charge

ε₀ is permittivity of space

And from the definition of flux

Φ = EA

The flux is electric field passing perpendicularly through the surface

Put the this Φ in equation(i)

EA =Q/ε₀

where Q(charge)=σA

EA=(σA)/ε₀

E=σ/ε₀

σ=ε₀E


=(8.85*10^(-12) )*(75.0*10^(3) )\\=6.6375*10^(-7) C/m^(2)\\=6.63*10^(2)nC/m^(2)

σ=6.6375×10²nC/m²

For (b) total charge Q

Q=σA


Q=(6.6375*10^(2) nC/m^(2) )(0.47m)^(2)\\ Q=1.47**10^(2)nC

User Marilena
by
5.6k points