The answer is -1, because protons are positive charges and electrons are negative charges. If you subtract them you would get you answer if -1
Answer:
I₂ = I₀ 1/4
Explanation:
For this exercise we use the definition of intensity which is the power per unit area
I = P / A
The emitted power is constant, so
P = I A
We can write this equation for the start and end point with index 2
I₀ A₀ = I₂ A₂
I₂ = I₀ A₀ / A₂
The spot area is the area of the circle
A₀ = π r₀²
We substitute
I₂ = I₀ r₀² / r₂²
It indicates that the radius of the spot is twice the initial radius
r₂ = 2 r₀
I₂ = I₀ (r₀ / 2 r₀)²
I₂ = I₀ 1/4
Answer:
The Questions are
a. Calculate the x-component of the electric field, in newtons per coulomb
b. Calculate the y-component of the electric field, in newtons per coulomb
c. Calculate the z-component of the electric field, in newtons per coulomb
d. Calculate the magnitude of the electric field, in newtons per coulomb.
Explanation:
Given that,
Φx= 85 N•m2/C. X direction
Φy= -85 N•m2/C. Y direction
Φz = 0. Z direction
Radius of loop =3cm=0.03m
Surface area of the circle is πr²
A=22/7×0.03²
A=0.00283m²
Flux is given as Φ=EA
a. Φx=ExA
Ex=Φx/A
Ex=85/0.00283
Ex=30035.33N/C
b. Ey=Φy/A
Ey=-85/0.00283
Ey=-30035.33N/C.
c. Ez=Φz/A
Ez, =0/A
Ez=0N/C
d. Magnitude of E.
E=√Ex²+Ey²+Ez²
E=√(30035.33)²+(-30035.33)²
E=42476.38N/C.