The electric flux on a surface of area A in a uniform electric field E is given by
φ = EA cos θ
where θ = the angle between the directions of E and a normal vector to the surface of the area.
See the diagram shown below.
When the electric field is perpendicular to the surface, then θ = 0°, and
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) = 1.984 x 10⁶ (N-m²)/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0
Answer:
(a) 1.984 x 10⁶ (N-m²)/C
(b) 0
Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N
, c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
They cannot form on a Transform Boundary
The basic structure would be:
Reactants → Products
Answer:
The correct statement is "The electric field is directed toward the electron and has a magnitude of 
".
Explanation:
According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by
- k is the Coulmob's constant.
The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.
Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.
The charge on the electron = -e.
The electric field due to an electron is given by
The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.
Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of ".
