Find out the surface charge density at the intersection of point X=3 {m} plane and X-axis, in the region of uniform line charge of 8 {nC}/{m} lying along the Z-axis in free space [16 Mar 2021 Shift 2]

Correct Answer is : 0.424nC m−20.424 {nC m}^{-2}0.424nC m−2

Correct Answer is : 0.424nC m−20.424 {nC m}^{-2}0.424nC m−2

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Electrostatics Part 2

Section: Physics

Question : 96 of 100

Marks: +1, -0

X=3 \text{m}

X

8 \text{nC}/\text{m}

Z

Solution:

Given,

Linear charge density,

\lambda=8 \text{nC}/\text{m}=8 \times 10^{-9} \text{C}/\text{m}

The relation between surface charge density and linear charge density can be given as

\;\frac{2 k \lambda}{r}=\;\frac{\sigma}{\varepsilon_{0}}

.....(i)

where,

k=

Coulomb's constant,

\lambda=

linear charge density,

\sigma=

surface charge density,

\varepsilon_{0}=

absolute electrical permittivity of free space

and

r=

distance.

Substituting the values in Eq. (i), we get

\sigma=\;\frac{2 k \lambda \varepsilon_{0}}{r}

=\;\frac{2 \times 9 \times 10^{9} \times 8 \times 10^{-9} \times 8.85 \times 10^{-12}}{3}

=0.424 \times 10^{-9} \text{Cm}^{-2}=0.424 \text{nCm}^{-2}

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