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aalyn [17]
2 years ago
6

If a dielectric material, such as teflon®, is placed between the plates of a parallel-plate capacitor without altering the struc

ture of the capacitor, how is the capacitance affected?.
Engineering
1 answer:
erastovalidia [21]2 years ago
4 0

The capacitance is affected is that the capacitance increases because of the insertion of Teflon.

<h3>What is capacitance?</h3>

A `is known to be a form of a device that is said to be used to save some amount of charges in any electrical circuit.

Note that the capacitor functions on the principle that the capacitance of a conductor is said to increases if an earthed conductor is  taken close to it.

Hence, The capacitance is affected is that the capacitance increases because of the insertion of Teflon.

See full question below

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structure of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected?

The electric field is not altered, because the structure remains unchanged.

The electric field becomes zero after the insertion of the Teflon®.

The electric field decreases because of the insertion of the Teflon®.

The electric field becomes infinite because of the insertion of the Teflon®.

The electric field increases because of the insertion of the Teflon®.

Learn more about  electric field  from

brainly.com/question/14781269

#SPJ1

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The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

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The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

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So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

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The drawings reflect what the project specifications are unlikely to cover

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