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3 years ago

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Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces. The magnitu

de of the surface charge density on either of the facing surfaces is 2.0 nc/m2. Determine the magnitude of the electric potential difference between the plates.

1 answer:

Pepsi [2]3 years ago

8 0

Answer:

18.1 V

Explanation:

The electric field between two parallel plates is given by the equation:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the charge surface density

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

For the plates in this problem,

$\sigma = 2.0 nC/m^2 = 2.0\cdot 10^{-9} C/m^2$

So, the magnitude of the electric field is

$E=\frac{2.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=226.0 V/m$

Now we can find the potential difference between the plates, which is given by

$\Delta V = E d$

where

d = 8.0 cm = 0.08 m is the separation between the plates

Substituting,

$\Delta V=(226.0)(0.08)=18.1 V$

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1 year ago

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

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Answer:

Explanation:

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3 years ago

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Hoochie [10]

Answer:

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Explanation:

Let $V(\text{Initial})$ and $P(\text{Initial})$ denote the volume and pressure of this gas before the compression.
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By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

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For this particular question:

$V(\text{initial}) = 1.0\; \rm m^3$ .
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Rearrange the equation to obtain:

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Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

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Also see attached file

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3 years ago