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3 years ago

Earth Science help needed

Physics

2 answers:

denis23 [38]3 years ago

6 0

D it is D the answer is D

Maurinko [17]3 years ago

5 0

Answer:

It is letter D. core, mantle crust

Explanation:

i got this from study island i got the same question

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A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just

Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

$E = \frac {\sigma }{2\epsilon _{0}}$

$E = \frac{9\times 10^{-6}}{2\times 8.854\times 10^{-12}}$

E = 5.08 x 10^5 N/C

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3 years ago

You have to choose one of these Kpop members for the purge, or you can group them into three. Who do you choose, what are their

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2 years ago

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A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and

Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

$q_{out}=q_{2}-q_{1}$

$q_{out}=-6.43-(-4.00)$

$q_{out}=-2.43\ \mu C$

The charge on the inner surface is q.

$q+(-2.43)=-6.43$

$q=-6.43+2.43= 4.00\ \mu C$

We need to calculate the electric field outside the shell

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}$

$E=33927618.73\ N/C$

$E=3.39\times10^{7}\ N/C$

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is $3.39\times10^{7}\ N/C$

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2 years ago

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3 years ago

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