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3 years ago

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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 di

rected 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

1 answer:

timurjin [86]3 years ago

6 0

Answer:

(a) Charge density σ=6.6375×10²nC/m²

(b) Total charge Q=1.47×10²nC

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

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

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