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Hitman42 [59]
3 years ago
6

Anyone good with scientific notations? I sure am not

Physics
2 answers:
defon3 years ago
7 0
B is the answer that I know of.
algol133 years ago
5 0
I'm pretty sure it is b but don't take my word for it
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Explain the molecular concept of thermal energy.
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A step-up transformer is connected to a generator that is delivering 141 V and 110 A. The ratio of the turns on the secondary to
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3 years ago
A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
Andreas93 [3]

Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

6 0
2 years ago
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