long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m3) distributed throughout its volume. Dete
rmine the magnitude of the electric field 15 cm from the axis of the cylinder.
1 answer:
Answer:
the magnitude of the electric field = 27.1058 V/m
≈ 27 V/m
Explanation:
given
R(radius enclosed) = 12 cm = 0.12 m
ρ(charge density) = 5.0 nC/m³ = 5.0 × 10⁻⁹C/m³
r(radius from the axis) = 15 cm = 0.15 m
using Gauss law, which states that the electric flux through any closed surface is directly proportional to the total electric charge enclosed by this surface.
attached is the calculation of the question, using Gauss law of electrostatics
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Explanation:
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Hence, this is the required solution.
Answer
The rate at which the magnetic field is changing is
Explanation
From the question we are told that
The electric field strength is
The radius is
The rate of change of the magnetic field is mathematically represented as
Where is change of a unit length
Where A is the area which is mathematically represented as
So
where L is the circumference of the circle which is mathematically represented as
So
substituting values
Answer:
1.6 s
Explanation:
To find the time in which the potential difference of the inductor reaches 24V you use the following formula:
V_o: initial voltage = 60V
R: resistance = 24-Ω
L: inductance = 42H
V_L: final voltage = 24 V
You first use properties of the logarithms to get time t, next, replace the values of the parameter:
hence, after 1.6s the inductor will have a potential difference of 24V