Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
The same speed of light, the fastest possible
darkness is just the absence of light therefore once light leaves darkness returns. making darkness have the same speed of light
Answer:
In D: 3J
Explanation:
Potential energy: Ep=mgh where m is the mass, h altitude.
In point A: h=20cm=0.2m
Epa=12=0.2×mg. Thus mg=12/0.2=60N
For point D: hd=5cm=0.05m
Epd=mg×0.05=60×0.05=3J
