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3 years ago

Helpppppppppp pppppplzzzzzzzzzzzzzzzzzzzzzzzzz

Physics

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

The answer is reflection.

alina1380 [7]3 years ago

4 0

D bend I hope that helped you

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1 year ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

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:

$E=\dfrac{\sigma}{2\epsilon_0}$

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,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

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

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3 years ago

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