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Question: A charged cylinder of radius 3 mm has surface density of charge 4 $\mu C/m^2$. placed in a medium of...

A charged cylinder of radius 3 mm has surface density of charge 4 μC/m2\mu C/m^2. placed in a medium of dielectric constant 6.28. The electric intensity at a point distance of 1.5 m from its axis is,

A

144 V/m

B

2.44 V/m

C

3 V/m

D

0.5 V/m

Answer

144 V/m

Explanation

Solution

For a uniformly charged cylinder:

  • Given:
    Radius, R=3 mm=0.003 mR = 3 \text{ mm} = 0.003\,m
    Surface charge density, σ=4 μC/m2=4×10−6 C/m2\sigma = 4\,\mu C/m^2 = 4\times10^{-6}\,C/m^2
    Dielectric constant, K=6.28K = 6.28
    Point distance, r=1.5 mr = 1.5\,m

  • Step 1: Calculate the linear charge density:

    λ=σ×(2πR)=4×10−6×(2π×0.003) C/m≈7.54×10−8 C/m.\lambda = \sigma \times (2\pi R) = 4\times10^{-6} \times (2\pi \times 0.003) \,C/m \approx 7.54\times10^{-8}\,C/m.

  • Step 2: Determine the permittivity of the medium:

    ϵ=Kϵ0=6.28×8.854×10−12 F/m≈5.56×10−11 F/m.\epsilon = K \epsilon_0 = 6.28 \times 8.854\times10^{-12}\,F/m \approx 5.56\times10^{-11}\,F/m.

  • Step 3: Use Gauss's Law for a line charge in a dielectric medium (for r>Rr > R):

    E=λ2πϵr.E = \frac{\lambda}{2\pi \epsilon r}.

  • Step 4: Substitute the values:

    E=7.54×10−82π×5.56×10−11×1.5.E = \frac{7.54\times10^{-8}}{2\pi \times 5.56\times10^{-11} \times 1.5}.

    Calculate the denominator:

    2π×5.56×10−11×1.5≈5.24×10−10.2\pi \times 5.56\times10^{-11} \times 1.5 \approx 5.24\times10^{-10}.

    Thus,

    E≈7.54×10−85.24×10−10≈144 V/m.E \approx \frac{7.54\times10^{-8}}{5.24\times10^{-10}} \approx 144\,V/m.