<span>that's a good question. It is a amorphous solid because it expands </span>
Answer:
1) Option D is correct.
The electric field inside a conductor is always zero.
2) Option A is correct.
The charge density inside the conductor is 0.
3) Charge density on the surface of the conductor at that point = η = -E ε₀
Explanation:
1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.
Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.
3) Surface Charge density = η = (q/A)
But electric field is given as
E = (-q/2πε₀r²)
q = -E (2πε₀r²)
η = (q/A) = -E (2πε₀r²)/A
For an elemental point on the surface,
A = 2πrl = 2πr²
So,
η = -E ε₀
Hope this Helps!!!
Answer: 207 W
Explanation: I assumed here that the object is moved vertically. If that is the case, the work done on the object is equal to its change in gravitational potential energy:
where
m = 47 kg is the mass of the object
g = 9.8 m/s^2 is the acceleration of gravity
is the change in height
Substituting,
Now we can calculate the power used, which is given by
Hope this helps I'm sorry if i'm wrong but I tried :(
