What are star system

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

padilas [110]

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A 4.0 cm diameter parallel plate capacitor has a 0.44 m m gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

5 0

3 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area

h = Altitude = 600 m

Electric flux through the top would be

$-110A$ (negative as the electric field is going into the volume)

At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0$

Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

The volume charge density is $1.475\times 10^{-13}\ C/m^3$

7 0

3 years ago

How do you measure the wavelength of a transverse wave?

DENIUS [597]

By looking at how wiggily the bar is lol

8 0

3 years ago

shusha [124]

Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.

Explanation:

4 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago