The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
<h3>Electric field at position 4 m</h3>
Electric field at a given distance is calculated as follows;
E = kq/r²
E₂ = (9 x 10⁹ x q)/(2²)
E₂ = 2.25 x 10⁹q
E₂ + E₀ = 0
2.25 x 10⁹q + 4 = 0
2.25 x 10⁹q = - 4
q = -4 / (2.25 x 10⁹)
q = -1.78 x 10⁻⁹
E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)
E₄ = - 1 N/C
|E₄| = 1 N/C
Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
The complete question is below:
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
Learn more about electric field here: brainly.com/question/14372859
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Answer:
10.85 kV
Explanation:
The maximum value of the emf (ε) produced can be calculated using Faraday's law equation:
(1)
<u>Where:</u>
N: is the turns of wire = 130 turns
Φ: is the magnetic flux = BAcos(θ)
B: is the magnetic field = 3.82 T
A: is the area of the coil = a*b = 0.746 m*0.249 m = 0.186 m²
θ: is the angle between the magnetic field lines and the normal to A = ωt = 2πft
f: is the frequency = 1120 rev/min = 18.7 rev/s
From equation (1) we have:
Therefore, the maximum value of the emf produced is 10.85 kV.
I hope it helps you!
the answer should be:
When the buoyant force is equal to the force of gravity
Answer:
Image will form at distance 22.22 cm behind the mirror
Explanation:
As we know that the mirror formula is given as
now we know that
object distance from mirror is
Focal length of the mirror is given as
now we have
The chemical formula for water, H2o means that each water molecule contains one oxygen atom and two hydrogen atoms. This is the formula for water which has a liquid form, a solid form as ice, and also a gaseous form as water vapor.
