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4 years ago

9

Two oppositely charged parallel plates are a fixed distance apart. Which phrase best describes the relationship between the elec

tric field intensity (E) between the plates and the potential difference (V) across the plates?

1 answer:

Ulleksa [173]4 years ago

3 0

Answer:

the electric field is proportional to the voltage divided by the separation of the plates,

Explanation:

This parallel plate configuration is widely used, as it creates a uniform electric field between the plates. Where the potential and field differences are related by

DV = E d

That is, the electric field is proportional to the voltage divided by the separation of the plates, in general all these configurations are called capacitors

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State three factors that determine the strength of electromagnetic

Sunny_sXe [5.5K]

Answer:

Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.Nov 28, 2020

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3 years ago

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When Pluto was classified as a planet it was known as a oddball planet why? Why is it less if an oddball now?

hichkok12 [17]

Answer:

a strange world that has baffled scientists ever since it was discovered in 1930. It is not the large gas giant that one might expect to find in the outer reaches of the solar system.

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3 years ago

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

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3 years ago

You were in a tog-o-war contest which your team won. Write a letter to your friend that tells why your team won the contest. Rem

ollegr [7]

Answer:

Imagine a tug-of-war between you and one

friend. If you are stronger, you apply more

force to the rope. You pull your friend across

the line, and you are the winner! If your friend

is stronger, he might pull you across the line.

Sometimes the forces are equal. Neither you

nor your friend moves across the line. The two

forces are balanced.

3

We say that the net force on an object is the combination of all the forces acting on it. To find the net force of

forces that are acting in the same direction, add them together. For example, if you pull on a box with a force of 25

newtons (N) while your friend pushes the box (in the same direction you are pulling) with a force of 30 N, the net

force applied to the box in that direction is 55 newtons.

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To find the net force of forces that are acting in opposite directions, subtract the smaller force from the larger

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the opposite direction, the net force on the rope is 5 newtons in your direction. You win!

Explanation:

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3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago