Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of rad

Question

Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical surfaces of radii 1.0 mm and 3.0 mm. Determine the magnitude of the electric field at a point which is 2.0 mm from the symmetry axis.

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Khánh Gia 1 week 2021-07-21T22:26:56+00:00 1 Answers 3 views 0

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    2021-07-21T22:28:30+00:00

    Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.

    Explanation:

    Given: Density = 80 nC/m^{3} (1 n = 10^{-9} m) = 80 \times 10^{-9} C/m^{2}

    r_{1} = 1.0 mm (1 mm = 0.001 m) = 0.001 m

    r_{2} = 3.0 mm = 0.003 m

    r = 2.0 mm = 0.002 m (from the symmetry axis)

    The charge per unit length of the cylinder is calculated as follows.

    \lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})

    Substitute the values into above formula as follows.

    \lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})\\= 80 \times 10^{-9} \times 3.14 \times [(0.003)^{2} - (0.001)^{2}]\\= 2.01 \times 10^{-12} C/m

    Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.

    E = \frac{\lambda}{2 \pi r \epsilon_{o}}

    Substitute the values into above formula as follows.

    E = \frac{\lambda}{2 \pi r \epsilon_{o}}\\= \frac{2.01 \times 10^{-12} C/m}{2 \times 3.14 \times 0.002 m \times 8.85 \times 10^{-12}}\\= 18.08 N/C

    Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.

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