Physics, 17.03.2020 04:11 1tzM3

The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates

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Answer from: torigirl4126

(a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

Explanation:

Given that,

Area = 8.50 cm²

Distance = 3.00 mm

Potential = 6.00 V

Distance without discharge = 8.00 mm

(a). We need to calculate the capacitance

Using formula of capacitance

$C_{1}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{1}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{3.00\times10^{-3}}$

$C_{1}=2.5075\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

Q=CV

Put the value into the formula

$Q=2.5075\times10^{-12}\times6.00$

$Q=1.5045\times10^{-11}\ C$

(b). We need to calculate the initial stored energy

Using formula of initial energy

$E_{i}=\dfrac{1}{2}\times CV^2$

$E_{i}=\dfrac{1}{2}\times2.5075\times10^{-12}\times36$

$E_{i}=4.5135\times10^{-11}\ J$

(c). We need to calculate the capacitance

Using formula of capacitance

$C_{2}=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C_{2}=\dfrac{8.85\times10^{-12}\times8.5\times10^{-4}}{2\times8.00\times10^{-3}}$

$C_{2}=9.403\times10^{-13}\ F$

We need to calculate the final stored energy

Using formula of initial energy

$E_{f}=\dfrac{1}{2}\times \dfrac{Q^2}{C}$

$E_{f}=\dfrac{1}{2}\times\dfrac{(1.5045\times10^{-11})^2}{9.403\times10^{-13}}$

$E_{f}=12.036\times10^{-11}\ J$

(d). We need to calculate the work done

Using formula of work done

$W=E_{f}-E_{i}$

Put the value in the formula

$W=12.036\times10^{-11}-4.5135\times10^{-11}$

$W=7.5225\times10^{-11}\ J$

Hence, (a). The charge is $1.5045\times10^{-11}\ C$

(b). The initial stored energy is $4.5135\times10^{-11}\ J$

(c). The final stored energy is $12.036\times10^{-11}\ J$

(d). The work required to separate the plates is $7.5225\times10^{-11}\ J$

Answer from: Quest

nec chapter 5 contains

c. information on special occupancy areas

nec chapter 6 covers

a. special equipment

which of the following rules contain the words shall or shall not and must be adhered to?

b. mandatory rules

which of the following rules identify actions that are allowed but not required and typically cover options or alternative methods?

c. optional rules

which of the following serves as the introduction to the nec and provides an overview of the book?

a. nec article 90

organizations that test products for safety are

c. nrtls

the nec is published by the

a national fire protection association

the nec is divided into

b. 9 chapters

the nec provides informative annexes for

c. informational materials and examples

two terms from nec article 100 that trainees should become familiar with are

b. labeled and listed

explanation:

Answer from: Quest

a is the best answer your welcome

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