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Vesnalui [34]
3 years ago
15

What are 13 types of electric charge?

Physics
1 answer:
Mamont248 [21]3 years ago
8 0
That's a really wonderful question, my favorite ever from this site, but you'll have to settle for a mediocre answer. I think the important proofs are experimental, because the whole theoretical framework within which things like this can be 'proved' is based on experimental results, not written on the sky. 

<span>We couldn't truthfully say that there were only two types of electrical charge unless we had some way of distinguishing between electrical charges and the 'charges' involved in other forces (gravity and the nuclear forces). Let's say that by electrical charges, we mean the sort of charges that cause forces at appreciable distances (that excludes the nuclear forces) and which can be transferred by rubbing different objects together (that pretty much excludes gravity). </span>

<span>When we say 'there are only two types of charge' what we mean is that charge can be represented by an ordinary real number, corresponding to a point on the real number line. Real numbers can be either positive or negative. For any positive, you can find a negative just big enough to cancel it and give zero. </span>

<span>What's a simple experimental version of this? You may have seen the experiment with a couple of gold leaves connected to each other. Whenever an electrical charge is deposited on the leaves, they push apart, because they share the same type of charge. You can get electrical charges to put on them from all sorts of static electricity- rubbing a rock on a wool rug, and touching the rock to the gold, etc. Now for all the many ways you can dump charge onto the gold, you can divide the whole collection into two batches, which we’ll call N and P. For any charge in batch "P", you can always get back to zero (where the leaves touch) by dribbling in bits of charge of type "N", and vice versa. So that's just like positive and negative numbers, and is the basic justification for representing the charges with positive and negative numbers. </span>

<span>You might ask (since you sound like a pretty profound asker) whether any other situation is even imaginable. Try this. Say there were some sort of charge that were represented not by numbers but by position in a plane. Say you had some "north" charge, and some "east" charge. There's no way of combining them to get back to zero- so they can't be like numbers of opposite sign. And there's no other type of charge that could be used by itself to cancel both of them- so they aren't like numbers of the same sign. There's no way to divide these planar charges into two distinct batches, where you can cancel any charge in one batch with the right amount of any charge from the other batch. So you know these charges can't be represented by real numbers. </span>

<span>In case that sounds imaginative, I can't claim credit. Nature has a lot of imagination. The 'color charges' of the strong nuclear force are pretty much like that. </span>
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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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The specific heat of a certain type of metal is 0.128 J/(g⋅∘C).0.128 J/(g⋅∘C). What is the final temperature if 305 J305 J of he
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Answer:

45.3°C

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = mC (T − T₀)

Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:

305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)

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