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eimsori [14]
3 years ago
11

A long, straight metal rod has a radius of 5.70 cm and a charge per unit length of 28.2 nC/m. Find the electric field at the fol

lowing distances from the axis of the rod, where distances are measured perpendicular to the rod's axis. (a) 3.10 cm magnitude N/C direction (b) 18.0 cm magnitude N/C direction(c) 180 cm magnitude N/C direction
Physics
1 answer:
tekilochka [14]3 years ago
5 0

Answer:

0

2817.43213 N/C

281.74321 N/C

Explanation:

The metal rod is conducting electricity. This means that the charges are being carried on the surface. Inside the rod there is no electric field. So, the electric field intensity is zero at a distance 3.1 cm.

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

r = Distance

\lambda = Charge per unit length = 28.2 nC/m

Electric field is given by

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 0.18}\\\Rightarrow E=2817.43213\ N/C

The electric field intensity is 2817.43213 N/C

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 1.8}\\\Rightarrow E=281.74321\ N/C

The electric field intensity is 281.74321 N/C

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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;

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length of the conductor, L = 50 m

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outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

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