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zhuklara [117]
3 years ago
15

How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant

is 9.0?
Physics
2 answers:
Nady [450]3 years ago
7 0
:<span>  </span><span>Under the assumption that a cell is made up of two concentric spheres you find the surface are of the inside sphere which will be your A. 

You already have your separation and dielectric constant so just use the formula you stated towards the end of your question and you get 8.93x10^-11 Farads which is about 89pF</span>
hichkok12 [17]3 years ago
3 0

The capacitance of the capacitor will be \boxed{8.9\times10^{-11}\text{ F}} or \boxed{89\text{ pF}}.

Explanation:

Given:

The diameter of the plates of the capacitor is 50\,\mu\text{m}.

The distance between the plates is 7\text{ nm}.

The dielectric constant of the medium is 9.0.

Concept:

The two conducting plates kept parallel to each other and placed at a particular distance from one another are considered to be the parallel plate capacitor.

The capacitance of the parallel plates capacitor formed by the two plates is given by the expression:

\boxed{C=\dfrac{K\epsilon_0A}{d}}

Here, C is the capacitance of the plates, K is the dielectric constant of the medium, \epsilon_0 is the permittivity of free space, A is the area of the plates and d is the distance between the plates.

The area of the plates is:

\begin{aligned}A&=\pi R^2\\&=3.14\times(25\times10^{-6})^2\\&=1.96\times10^{-9}\text{ m}^2\end{aligned}

Substitute the values in equation of capacitance.

\begin{aligned}C&=\dfrac{9.0\times8.85\times10^{-12}\times1.96\times10^{-9}}{7.0\times10^{-9}}\\&=8.9\times10^{-11}\text{ F}\\&=89\text{ pF}\end{aligned}

Thus, The capacitance of the capacitor will be \boxed{8.9\times10^{-11}\text{ F}} or \boxed{89\text{ pF}}.

Learn More:

1. Change in momentum due to its collision: brainly.com/question/9484203  

2. Type of mirror used by dentist: brainly.com/question/997618  

3. Energy density of the energy stored between the plates of capacitor brainly.com/question/9617400

Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Capacitor

Keywords:

capacitor, capacitance, electric, parallel, plates, particular distance, dielectric constant, permittivity, area of the plates.

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Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

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Explanation:

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a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

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E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

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The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

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