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Physics / Physics with MasteringPhysics 4 / Chapter 20 / Problem 55P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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IP A parallel-plate capacitor has plates with an area of

Chapter 20, Problem 55P

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QUESTION:

Problem 55P

IP A parallel-plate capacitor has plates with an area of 0.012 m2 and a separation of 0.88 mm. The space between the plates is filled with a dielectric whose dielectric constant is 2.0. (a) What is the potential difference between the plates when the charge on the capacitor plates is 4.7 μC? (b) Will your answer to part (a) increase, decrease, or stay the same if the dielectric constant is increased? Explain, (c) Calculate the potential difference for the case where the dielectric constant is 4.0.

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QUESTION:

Problem 55P

IP A parallel-plate capacitor has plates with an area of 0.012 m2 and a separation of 0.88 mm. The space between the plates is filled with a dielectric whose dielectric constant is 2.0. (a) What is the potential difference between the plates when the charge on the capacitor plates is 4.7 μC? (b) Will your answer to part (a) increase, decrease, or stay the same if the dielectric constant is increased? Explain, (c) Calculate the potential difference for the case where the dielectric constant is 4.0.

ANSWER:

Step 1 of 3:-

a)

Here we have to calculate the potential difference between the plates of the capacitor.

The area of the plate is,

$A=0.012\ {m}^{2}$ .

The separation between the plates is,

$d=0.88\ mm=0.88\times{}1{0}^{-3}\ m$ .

The charge on the plates is,

$Q=4.7\ \mu{}C=4.7\times{}1{0}^{-6}\ C$ .

The dielectric constant of the dielectric present is,

$k=\epsilon{}/{\epsilon{}}_{0}=2$

$\Rightarrow{}\epsilon{}=k{\epsilon{}}_{0}$

$=2\times{}8.85\times{}1{0}^{-12}$

$=1.17\times{}1{0}^{-11}\ F/m$ .

The potential difference can be calculated by the following steps given below.

$C=\frac{Q}{V}=\frac{\epsilon{}A}{d}$

$\Rightarrow{}V=\frac{Qd}{\epsilon{}A}$

$=\frac{4.7\times{}1{0}^{-6}\ C\times{}0.88\times{}1{0}^{-3}\ m}{1.17\times{}1{0}^{-11}\ F/m\times{}0.012\ {m}^{2}}$

$=19472.7\ volts$

$=19.472\ KV$ .

So, the potential difference is 19.472 KV.

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