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Softa [21]
3 years ago
10

Ask Your Teacher When a potential difference of 160 V is applied to the plates of a parallel-plate capacitor, the plates carry a

surface charge density of 26.0 nC/cm2. What is the spacing between the plates?
Physics
1 answer:
Morgarella [4.7K]3 years ago
7 0

The spacing between the plates is 5.4\cdot 10^{-6} m

Explanation:

The electric field in the spacing between the plates of a parallel-plate capacitor is uniform and it is given by

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

where

\sigma is the surface charge density

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

We also know that the relationship between electric field and potential difference across a parallel-plate capacitor is

V=Ed

where

V is the potential difference

d is the spacing between the plates

Combining the two equations we get

V=\frac{\sigma d}{\epsilon_0}

And given that here we have:

V = 160 V

\sigma = 26.0 nC/cm^2 = 26.0\cdot 10^{-9} C/cm^2=26.0\cdot 10^{-5} C/m^2

We can solve the equation for d, the spacing between the plates:

d=\frac{V \epsilon_0}{\sigma}=\frac{(160)(8.85\cdot 10^{-12})}{26.0\cdot 10^{-5}}=5.4\cdot 10^{-6} m

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

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Explanation:

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(40 x π ) / 180 = 410 / d

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3 years ago
a hot iron bar is placed 100ml 22C water. the water temperature rises to 32C. how much heat did the water gain and how much heat
Sergeu [11.5K]

Answer:

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Explanation:

Given: Hot iron bar is placed 100ml 22C water, the water temperature rises to 32C

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Formula:Q = change T x C x M

Solve:

<u>How much heat water gained</u>

Initial heat = 22, then rose to 32. To find how much heat the water gained, simply subtract the current heat by the initial heat.

                                              32 - 22 = 10

The water gained 10 amounts of heat.

<u>How much heat Iron lost</u>

Current heat = 32, then dropped to 22. To find how much heat the Iron lost, simply subtract the initial heat by the current heat.

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Q = mcDt

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Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

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siniylev [52]

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