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3 years ago

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A 2.98 nF parallel-plate capacitor is charged to an initial potential difference of 49 V and then isolated. The dielectric mater

ial between the plates has a dielectric constant of 3.1. What is the potential difference of the capacitor after the dielectric is withdrawn? Answer in units of V.

2 answers:

I am Lyosha [343]3 years ago

8 0

Answer:

Potential difference will be 151.9 volt

Explanation:

We have given capacitance of the capacitor $C=2.98nF=2.98\times 10^{-9}F$

Voltage V = 49 Volt

Dielectric constant K = 3.1

We have to find the potential difference

We know that when a dielectric medium is introduced then p[otential difference is increases by k times

As the dielectric constant k = 3.1

So potential difference will be = 3.1×49 = 151.9 volt

docker41 [41]3 years ago

6 0

Answer:151.9 V

Explanation:

Given

Capacitance (C)=2.98 nF

Potential difference=49 V

dielectric strength k=3.1

Charge remains same after the removal of dielectric

thus $Q_1=Q_2$

$Q_1=2.98\times 49$

Now dielectric is removed so capacitance decreases to $\frac{C}{k}$

$Q_2=\frac{2.98}{3.1}\times V$

$V=49\times 3.1=151.9 V$

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Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

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$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

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The general equation of travelling wave is given by :

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Fiesta28 [93]

Answer:

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Explanation:

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4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

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1. equation 1 is equal to:

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3 years ago