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.

Electrostatics PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q58 with Solution

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Correct answer: A

A uniform line charge of density

\lambda = 8\,\text{nC/m}

lies along the $z$ -axis in free space.

We need the surface charge density at the point where the plane $x=3\,\text{m}$ intersects the $x$ -axis.

The plane $x=3$ is a Gaussian surface boundary. The required point on this plane and on the $x$ -axis is:

(3,0,0)

For a field in free space, the surface charge density on an imaginary surface is related to electric flux density by

\rho_s = \mathbf{D}\cdot \hat{n}

where $\hat{n}$ is the unit normal to the surface.

For an infinite line charge,

\mathbf{D} = \frac{\lambda}{2\pi \rho}\,\hat{\rho}

where $\rho$ is the perpendicular distance from the line charge.

Since the line charge is along the $z$ -axis, the perpendicular distance of the point $(3,0,0)$ from the $z$ -axis is

\rho = \sqrt{x^2+y^2} = \sqrt{3^2+0^2}=3\,\text{m}

Thus,

D = \frac{8\times 10^{-9}}{2\pi(3)}

D = \frac{8\times 10^{-9}}{6\pi} approx 4.24\times 10^{-10}\,\text{C/m}^2

The plane $x=3$ has unit normal along $+\hat{x}$ .

At the point $(3,0,0)$ , the radial direction from the $z$ -axis is also along $+\hat{x}$ . So,

\mathbf{D}\cdot \hat{n} = D

Hence the surface charge density is

\rho_s = 4.24\times 10^{-10}\,\text{C/m}^2

Convert to nC/m $^2$ :

\rho_s = 0.424\,\text{nC/m}^2

The correct option is:

\boxed{\text{A: }0.424\,\text{nC m}^{-2}}