Physics, 25.03.2020 22:32 amyeileen

Two identical air-filled parallel-plate capacitors C1 and C2, each with capacitance C, are connected in series to a battery that has voltage V. While the two capacitors remain connected to the battery, a dielectric with dielectric constant K>1 is inserted between the plates of one of the capacitors, completely filling the space between them. Let U0 be the total energy stored in the two capacitors without the dielectric and U be the total energy stored after the dielectric is inserted.
a) What is the ratio U/U0?
b) Does the total stored energy increase, decrease, or stay the same after the dielectric is inserted?

Answers: 3

Show answers

Answers

Answer from: astultz309459

a) The ratio is $\frac{U}{U_{0}} =\frac{2}{(1+\frac{1}{\kappa})}$ .

b) The total stored energy increases after the dielectric is inserted.

Explanation:

Before the dielectric is inserted:

We have two parallel plate capacitors ( $C_{1}$ and $C_{2}$ ), each with a capacitance C, connected in series to a battery that has voltage V.

The two capacitors can be replaced by an equivalent one. The equivalent capacitance for two capacitors in series is

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{2}{C}$

The potential differences across the two capacitors are:

$\Delta V_{1}=\frac{Q}{C_{1}}=\frac{Q}{C}$ and $\Delta V_{2}=\frac{Q}{C_{2}}=\frac{Q}{C}$

The total potential difference $\Delta V$ is the sum of the potential differences across each capacitor:

$\Delta V=\Delta V_{1}+ \Delta V_{2}=2\frac{Q}{C}$

The initial stored energy $U_{0}$ is:

$U_{0}=\frac{1}{2}C_{eq} \Delta V^{2}=\frac{1}{2}\frac{C}{2}(2\frac{Q}{C})^{2} \ \Longrightarrow\ U_{0}=\frac{Q^{2}}{C}$

Using $\Delta V=V$ instead we have that $U_{0}=\frac{1}{4}CV^{2}$ .

After the dielectric is inserted:

A dielectric with $\kappa 1$ is inserted between the plates of one of the capacitors while the two capacitors are connected to the battery. The total potential difference reamins the same. If Q is the charge in the plates without the dielectric, in the capacitor with the dielectric the charge increases to $\kappa Q$ and the capacitance to $\kappa C$ .

Now we have that the equivalent capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{\kappa C}=\frac{1}{C} (1+\frac{1}{\kappa})$

So the stored energy after the dielectric is inserted becomes:

$U=\frac{1}{2}C_{eq}\Delta V^{2}=\frac{1}{2} \frac{C}{(1+\frac{1}{\kappa})}V^{2} \Longrightarrow\ U=\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}$

So, the ratio $\frac{U}{U_{0}}$ is:

$\frac{U}{U_{0}}=\frac{\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}}{\frac{1}{4}CV^{2}} \Longrightarrow\ \frac{U}{U_{0}}=\frac{2}{(1+\frac{1}{\kappa})}$

Because $\frac{U}{U_{0}} 1$ if $\kappa 1$ this means that the total stored energy increases after the dielectric is inserted.

Answer from: robertotugalanp1wlgs

(a) The required ratio of U/U0 for the given sets of capacitors is 2 / (1 + 1/k).

(b) The total stored energy increases after the dielectric is inserted.

Given data:

The capacitance of air-filled parallel plate capacitors are, C1 and C2.

The net capacitance is, C.

The battery voltage is, V.

The total energy stored in the two capacitors without dielectric is, U0.

The total energy stored with dielectric is, U.

The given problem is based on the concepts and fundamentals of parallel plate capacitors.

(a)

Before inserting the dielectric we have equivalent capacitance as,

$\dfrac{2}{C}=\dfrac{1}{C1}+\dfrac{1}{C2}$

And the potential across the two capacitors are,

V1 = Q/C1

V1 = Q/C

And,

V2 = Q/C2

V2 = Q/C

Total potential across the each capacitor is,

V = V1 + V2

V = Q/C + Q/C

V = 2Q/C

the energy stored in the capacitor is,

U0 = 1/2 C × V²

U0 = 1/2 (C/2) × (2Q/C)²

U0 = 1/2 (C/2) × 4Q²/C²

U0 = Q²/C ......................................................(1)

Now after inserting the dielectric, the capacitance becomes,

$\dfrac{1}{C'}=\dfrac{1}{C}+\dfrac{1}{ k C}\\\\\dfrac{1}{C'}=\dfrac{1}{C} \times (1+\dfrac{1}{ k })$

the stored energy after the dielectric is inserted U becomes:

$U = \dfrac{1}{2} \times \dfrac{C \times V^{2}}{(1+1/k)}$ ...............................................(2)

Taking the ratio of equation (2) and (1) as,

U/U0 = 2 / (1 + 1/k)

Thus, we can conclude that the required ratio of U/U0 for the given sets of capacitors is 2 / (1 + 1/k).

(b)

From the previous part we obtained that,

U/U0 = 2 / (1 + 1/k)

Clearly, U/U0 > 1 and also it is given that k > 1. Which significantly means that the total stored energy increases after the dielectric is inserted.

Thus, we can conclude that the total stored energy increases after the dielectric is inserted.

Learn more about the dielectrics here:

Answer from: Quest

Fitness is the ability to perform only very light levels of physical activity.

Answer from: Quest

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2 where t1=temperature of cold reservoir ( ° k) = 135 t2=temperature of hot reservoir ( ° k) = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58% note: this is the theoretical maximum efficiency of a perfect heat engine. real-life engines do not achieve this value.

Another question on Physics

Physics, 22.06.2019 09:30

Gasoline comes from petroleum, which is made from ancient living things. petroleum, therefore, contains a type of energy. a. heat b. nuclear c. biological d. chemical potential

Answers: 2

Answer

Physics, 22.06.2019 12:00

If the new moon happens on january 15th, what shape will it be on february 6th?

Answers: 1

Answer

Physics, 22.06.2019 13:40

Dao makes a table to identify the variables used in the equations for centripetal acceleration. what quantities belong in cells x and y?

Answers: 2

Answer

Physics, 22.06.2019 17:20

The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by b. what is the magnitude of the magnetic field at twice the distance from the conductor? a. at twice the distance, the magnitude of the field is 4b.b. at twice the distance, the magnitude of the field is 2b.c. at twice the distance, the magnitude of the field is b/2.d. at twice the distance, the magnitude of the field is b/4.e. at twice the distance, the magnitude of the field remains equal to b.

Answers: 3

Answer

You know the right answer?

Two identical air-filled parallel-plate capacitors C1 and C2, each with capacitance C, are connected...

Questions