Need help pls again fast

A 5.00-pf parallel-plate air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to p

r-ruslan [8.4K]

A 5 <em>pF</em> parallel-plate air-filled capacitor, with a potential difference equal to 100 Volts and an electric field of 10.000 Volts per meter, must have a plate separation of 10 milimeters, and plate's area equal to 0.0056 meters squared.

<h3>Further explanation</h3>

Capacitors are electrical components use to store electrical energy in the form of potential difference. This is done by putting 2 plates of a conductive material very close to each other, but without getting them into contact. The idea is that, when you charge one plate with (for example) positive charges, the other plate will "feel" this and will charge itself with negative charges (due to the attraction between charges of opposite signs).

The trick is that, even though the charges in the two plates attract each other, they can't come into contact since the 2 plates are separated by a certain distance. Now let's suppose there is only air between the capacitor's plates, we know that air is not a conductive material, however if you charge both plates just enough, you'll be able to make air conduct electricity and so charges from one plate will go to the other and equilibrium will be achieved.

This is the same thing that happens when a lightning strikes the ground. Imagine that the Earth is a gigantic capacitor, with the clouds being one plate and the ground being the other plate. When the clouds are charged enough, air is forced to conduct electricity towards the ground (which is the lightning we see), until equilibrium is achieved. This is the reason why, in Physics, we say that everything is a conductive material if you force it just enough.

Back to our problem... We can compute the separation of the capacitor's plates with the following equation:

$V = E\cdot d$

Which says that the Voltage <em>V</em> is equal to the product of the Electric field <em>E</em> and the distance between the plates <em>d</em>. This equation is valid if the electric field across the capacitor is constant (which for small magnitudes of <em>d</em> it is). So solving for the distance we get:

$d= \frac{V}{E} = \frac{100}{10000} \frac{V}{V/m} = 0.01 m = 10 mm$

Where we can see that the separation of plates is 10 millimeters. Now for the area between the plates we can use this other formula:

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

Where $\epsilon$ is the air's permittivity which has a value of around $8.86 \cdot 10^{-12} \frac{F}{m}$ . By solving for the plate's area, we can find:

$A= \frac{C \cdot d}{\epsilon} = 0.0056 m^2 =$

Which corresponds to a plate's radius of 42.4 millimeters.

<h3>Learn more</h3>

What is electricity: brainly.com/question/9194793
Relation between elementary particles and charge: brainly.com/question/12502147
Another capacitor problem: brainly.com/question/3203280

<h3>Keywords</h3>

Capacitor, Electric charge, Electric field, Voltage

4 0

3 years ago

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Which of these is NOT a physical property?

m_a_m_a [10]

I believe the answer is chemical reactivity because: Characteristics such as melting point, boiling point, density, solubility, color, odor, etc. are physical properties.

8 0

2 years ago

A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize

postnew [5]

Answer:

r=15.53 nm

$F=9.57\times 10^{-13}N$

Explanation:

Lets take electron is in between iron and uranium

Charge on electron $q_1= -1.602\times 10^{-19}C$

Charge on iron $q_2= 2\times 1.602\times 10^{-19}C$

Charge on uranium $q_3= 1.602\times 10^{-19}C$

We know that force between two charge

$F=K\dfrac{q_1 q_2}{r^2}$

$K=9\times 10^9\dfrac{N-m^2}{c^2}$

For equilibrium force between electron and iron should be force between electron and uranium

Lets take distance between electron and uranium is r so distance between electron and iron will be 37.5-r nm

Now by balancing the force

$K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}$

$K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}$

$q_2= 2\timesq_1,q_3=q_1$

$\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}$

So r=15.53 nm

So force

$F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}$

$F=9.57\times 10^{-13}N$

7 0

3 years ago

Which of the following has mechanical energy?

KATRIN_1 [288]

Mechanical energy can have mechanical systems. The only mechanical system in the list is the compressed spring. A car battery and a glowing incandescent lightbulb have electrical energy, a nucleus of atom has potential (internal) energy.

5 0

3 years ago

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You are travelling at 60.0 mph on the Grand Central Parkway near exit 10 where it is (nearly) parallel to the Long Island Expres

Romashka-Z-Leto [24]

Answer:

a)Vr= 5 mph

b)Vr= 115 mph

Explanation:

Lets

your velocity ,u= 60 mph

velocity of truck ,v= 55 mph

When object are moving in opposite direction the relative velocity = u +v When object are moving in opposite direction the relative velocity = u - v

a)

Truck is moving in the same direction ,so the relative velocity

Vr= = u - v

Vr= 60 - 55

Vr= 5 mph

b)

Truck is moving in the opposite direction ,so the relative velocity

Vr= = u+v

Vr= 60 + 55

Vr= 115 mph

4 0

3 years ago