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4 years ago

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

Physics

1 answer:

padilas [110]4 years ago

5 0

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

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3 years ago

Read 2 more answers

What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^

melisa1 [442]

Frequency: $1.27\cdot 10^8 Hz$

Explanation:

The force experienced by an electron in a magnetic field is

$F=qvB$

where

$q=1.6\cdot 10^{-19}C$ is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

r is the radius of the orbit

$m=9.31\cdot 10^{-31}kg$ is the mass of the electron

Re-arranging the equation,

$\frac{v}{r}=\frac{qB}{m}$ (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

$v=\frac{2\pi r}{T}$

Substituting into (1),

$\frac{2\pi}{T}=\frac{qB}{m}$

We also know that 1/T is equal to the frequency f, so

$f=\frac{qB}{2\pi m}$

In this problem,

$B=4.62\cdot 10^{-3}T$

Therefore, the frequency of the electrons is

$f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz$

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3 years ago

A child pushes a 100 kg refrigerator with a force of 50 N, but the refrigerator does not move. Suppose the coefficient of static

faust18 [17]

Answer:

50 N

Explanation:

Since the refrigerator doesn’t move, that means the force of friction equals the amount of force the child exerts on the fridge. If the friction force were greater than the force by the child, the fridge would start accelerating towards the child. If it were less than the force the child exerted, the fridge would start accelerating away from the child. Therefore, the net force must be 0, in this case, the friction force is equal to the force the child exerted, for it to stay at rest (as Newton’s First Law stated).

I hope this helps! :)

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3 years ago

A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6

Natasha2012 [34]

Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

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3 years ago