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3 years ago

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Addition of a metal slab of thickness "a" between the plates of a parallel plate capacitor of plate separation "d" is equivalent

to introducing a dielectric with dielectric constant "K" between the plates. Find an expression for "K" in terms of "a" and "d"?

1 answer:

Crazy boy [7]3 years ago

3 0

Answer:

$K = \frac{d}{d+a}$

Explanation:

The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

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

Where A = Area of the plate

d = distance between the plates

$\epsilon =$ dielectric constant

Case 1:

When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a

Therefore the capacitance of the capacitor becomes:

$C = \frac{\epsilon A}{d + a}$ .......................(1)

Case 2:

Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:

$C = \frac{K\epsilon A}{d}$ .........................(2)

Equating (1) and (2)

$\frac{K\epsilon A}{d} = \frac{\epsilon A}{d+a}\\\frac{K}{d} = \frac{1}{d+a} \\K = \frac{d}{d+a}$

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Read 2 more answers

a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature

igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber $V=1 m^3$

Initial Temperature $T_1=300 K$

Final Temperature $T_2=400 K$

Heat Supplied $Q=10 J$

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

$\Delta U=Q-W$

as the volume is fixed therefore work

$W=\int PdV=0$

thus $\Delta U=mc_v\Delta T=Q$

$c_v$ for mono-atomic gas is $12.471 J/K-mol$

$n\times 12.471\times (400-300)=10$

$n=0.008018 mol$

and 1 mole contains $6.022\times 10^{23} molecules$

thus No of molecules $=0.008018\times 6.022\times 10^{23}$

No of molecules $=4.82\times 10^{21} molecules$

3 0

3 years ago

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

REY [17]

Answer: $85.46\ kJ$

Explanation:

Given

Volume of air $V=105\ m^3$

Temperature of air $T=305\ K$

Increase in temperature $\Delta T=0.7^{\circ}C$

Specific heat for diatomic gas is $C_p=\dfrac{7R}{2}$

Energy required to increase the temperature is

$\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]$

Insert the values

$\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ$

6 0

3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

3 years ago