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Dima020 [189]
3 years ago
8

What the answer for this ? :/

Mathematics
1 answer:
goblinko [34]3 years ago
3 0
Hello there! I can help you! In order to answer those questions, we need to plug in the values, based off of the variable.

g. Okay. We are solving b - 10. b = -8. When you subtract something from a negative number, the number is even lower. Let's add the numbers first and then put in the negative symbol. 8 + 10 is 18. Put the negative sign and you get -18. The difference is -18.

h. Now, we solve a - b. a = 5 and b = -8. Because we are subtracting a negative number from a positive, we have to add both numbers, which means the number gets bigger. Ignore the negative sign and add. 5 + 8 is 13. There. The sum is 13.

i. The problem is c - a. c = -9 and a = 5. So as explained on problem G, for this problem, ignore the negative symbol and add. 9 + 5 is 14. Plug in the negative sign to get -14. There. The difference is -14.
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Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

The inner surface: Si=4\pi (0.06m)^{2} = 0.04 m^{2}

The outer surface: S=4\pi (0.08m)^{2}=0.08m^{2}

Now we can calculate the charges,

Inner charge: Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC

Outer charge: Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC

Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

Outer field:  Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

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