At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Fi
1 answer:
Answer:
a) Magnetic field strength, B = 2.397 * 10⁻⁷ T
b) Total energy density, U = 4.58 * 10⁻⁸ J/m³
c) Power flow per unit area, S = 13.71 W/m²
Explanation:
a) Electric field strength, E = 71.9 V/m
The relationship between the Electric field strength and the magnetic field strength in vacuum is:
E = Bc where c = 3.0 * 10⁸ m/s
71.9 = B * 3.0 * 10⁸
B = 71.9 / (3.0 * 10⁸)
B = 23.97 * 10⁻⁸
B = 2.397 * 10⁻⁷ T
b) Total Energy Density:
c)Power flow per unit area
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Explanation:
It is given that,
Speed, v₁ = 7.7 m/s
We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :
So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) =
V(r) - V(0) = ₀
V(r) = ₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) =
V(r) - V(0) =
V(r) =
V(r) =
V(r)
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V