Help ASAP <br> Just answer the first question for me please!

We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

8 0

3 years ago

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago

Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

Alex73 [517]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.320 m2. at

AlexFokin [52]

Remember, half of the energy in an EM wave is in the E field, the rest is in the B field. Thus, multiply E field energy by 2.
To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J

5 0

3 years ago

In optics, a diffraction ___________ is an optical component with a periodic structure, which splits and diffracts light into se

galina1969 [7]

In optics, a diffraction grating is an optical component with a periodic structure, which splits and diffracts light into several beams travelling in different directions.

4 0

3 years ago

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Help ASAP Just answer the first question for me please!