Astar begins as a blank a large cloud of gas and dust
1 answer:
You might be interested in
Answer:
the charge generated in the circuit is 240 C.
Explanation:
Given;
current flowing in the circuit, I = 2A
time of current flow, t = 2 minutes = 2 x 60s = 120 s
The current flowing through a given circuit is defined as the quantity of charge flowing through the circuit in a given time.
where;
Q is the charge flowing in the circuit
Q = 2 x 120
Q = 240 C
Therefore, the charge generated in the circuit is 240 C.
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 = /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
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
