A caterpillar tries to climb straight up a wall two meters high, but for every 2 cm up it climbs, it slides down 1 cm. Eventuall
1 answer:
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Answer:
Explanation:
m = Mass attached to spring = 14 kg
g = Acceleration due to gravity =
x = Displacement of spring =
k = Spring constant
The force balance of the system is given by
The spring constant for that spring is .
Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
The magnitude of the electric field is now given as;
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m