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Question

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

The correct answer is
$5\hat{c}$

Understanding Electric Flux Density for an Infinite Charge Sheet

The problem asks us to find the electric flux density, denoted by $\vec{D}$, at a specific point $P(0, 0, 5)$ caused by an infinite sheet of uniform charge lying on the $z=0$ plane.

  • Surface charge density ($\rho_s$) = $10 \text{ C/m}^2$.
  • The sheet is located at the $z=0$ plane.
  • The point $P$ is at $(0, 0, 5)$, which is directly above the sheet along the positive z-axis.
  • The medium has a relative permittivity ($\epsilon_r$) of 10.

Calculating Electric Flux Density ($\vec{D}$)

For an infinite sheet of charge with uniform surface charge density $\rho_s$, the electric flux density $\vec{D}$ is constant everywhere (except on the sheet itself) and is directed perpendicularly away from the sheet if $\rho_s$ is positive.

The formula for the electric flux density due to an infinite sheet of charge is:

$ \vec{D} = \frac{\rho_s}{2} \hat{n} $

Where:

  • $\rho_s$ is the surface charge density.
  • $\hat{n}$ is the unit normal vector pointing perpendicularly away from the sheet.

In this specific problem:

  • $\rho_s = 10 \text{ C/m}^2$.
  • The sheet lies on the $z=0$ plane. The point $P(0, 0, 5)$ is located at $z=5$, which is in the region $z > 0$. The unit normal vector pointing away from the sheet ($z=0$) into the region $z > 0$ is the positive z-direction unit vector, $\hat{n} = \hat{c}$.

Substituting the values into the formula:

$ \vec{D} = \frac{10 \text{ C/m}^2}{2} \hat{c} $ $ \vec{D} = 5 \hat{c} \text{ C/m}^2 $

The relative permittivity ($\epsilon_r = 10$) affects the electric field intensity ($\vec{E} = \frac{\vec{D}}{\epsilon_r \epsilon_0}$), but it does not change the formula for the electric flux density $\vec{D}$ itself, which depends directly on the surface charge density and geometry.

Final Result Analysis

The calculated electric flux density $\vec{D}$ at point $P(0, 0, 5)$ is $5 \hat{c} \text{ C/m}^2$. This matches one of the given options.

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Important Questions from Electrostatics

  1. A uniform ring charge of radius $R$ carries a total charge $Q$. Which one of the following options correctly quantifies the magnitude of the force on a point charge of strength $q$ kept at the center of the ring? 

    ($\epsilon$ is the permittivity of the medium)

  2. For a uniform line charge of 8 nc/m lying along the z axis, the electric field at a radius of 3 m from the uniform line is given by :

  3. The current in a coil of inductance 5 H changes linearly from 10 A to 2 A in t seconds. If the induced emf is 40 V, what is the time t?

  4. What are the materials which exhibit electric polarization even in the absence of an applied electrical field?

  5. What is the relative permittivity of slate dielectric?

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