Question

An infinite sheet of uniform charge \( \rho_s = 10\, {C/m}^2 \) is placed on the \( z = 0 \) plane. The medium surrounding the sheet has a relative permittivity of 10 . The electric flux density, in C/m\(^2\), at a point \( P(0, 0, 5) \), is: Note: \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors along the \( x, y, z \) directions, respectively.

• A. \( 5\, \hat{c} \)
• B. \( 0.25\, \hat{c} \)
• C. \( 10\, \hat{c} \)
• D. \( 0.5\, \hat{c} \)

Hint:

For an infinite sheet of charge, the electric flux density is calculated as \( \vec{D} = \frac{\rho_s}{2} \hat{n} \), which does not depend on the relative permittivity of the medium.

Answer 1

The Correct Option is A

Step 1: Use the formula for electric flux density due to an infinite sheet of charge.
For an infinite sheet of charge with surface charge density \( \rho_s \), the electric flux density \( \vec{D} \) is given by: \[ \vec{D} = \frac{\rho_s}{2} \hat{n} \] above the sheet, and \[ \vec{D} = -\frac{\rho_s}{2} \hat{n} \] below the sheet, where \( \hat{n} \) is the normal vector to the plane. Since the sheet lies in the \( z = 0 \) plane and point \( P(0,0,5) \) lies above it, the electric flux density at point \( P \) is: \[ \vec{D} = \frac{10}{2} \hat{c} = 5\, \hat{c} \quad {C/m}^2 \] Step 2: Understanding the medium and relative permittivity.
The electric flux density \( \vec{D} \) is independent of the medium’s relative permittivity \( \varepsilon_r \). Hence, the relative permittivity of \( 10 \) does not affect \( \vec{D} \). Thus, the final electric flux density at point \( P(0,0,5) \) is: \[ \vec{D} = 5\, \hat{c} \quad {C/m}^2 \] Conclusion:
Therefore, the correct answer is: \[ \boxed{A} \, 5\, \hat{c} \]