The expression for electric field intensity at a point which is outside a uniformly charged thin plane sheet is (d is the distance of point from plane sheet)
A. independent of $d$ $C{{m}^{-2}}$
B. directly proportional to $\sqrt{d}$
C. directly proportional to $d$
D. directly proportional to $\dfrac{1}{\sqrt{d}}$
Answer
65602602602602602312501250125012501250.7k+ views
Hint: Here we have to draw a cylindrical Gaussian surface.
The electric field intensity at a point is given by the formula,
$E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$
Where $E$ be the electric field vector, $\sigma $ is the electric charge density of the plane sheet and ${{\varepsilon }_{0}}$ is the specific inductive capacity or the relative permittivity of the thin plane sheet. And then find the relation between the terms.
Complete step-by-step answer:
First of all let us take a look at what actually the electric field intensity means. The electric field Intensity at a point is given as the force felt by a unit positive charge kept at a particular point. Electric Field Intensity is basically a vector quantity. It is denoted by $E$. It can be written in a formula like,
$E=\dfrac{F}{q}$
For an infinite sheet of charge, the electric field is found to be perpendicular to the surface. Because of this, only the ends of a cylindrical Gaussian surface will be contributing to the electric flux.
Here in this question, it is mentioned that the sheet is plane uniformly charged. Therefore the electric field is given by the equation,
$E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$
By looking at this equation we are able to find that the electric field intensity is not having any sort of connection with the distance of the point from the plain sheet. Therefore the correct answer is option A.
So, the correct answer is “Option A”.
Note: Surface charge density denoted as $\sigma $, calculated in coulombs per square meter, $C{{m}^{-2}}$ is an amount of charge per unit area at any point on a surface charge distribution, in the case of a two dimensional surface. Since electric charge will be either positive or negative, charge density can be expressed as either positive or negative values.
The electric field intensity at a point is given by the formula,
$E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$
Where $E$ be the electric field vector, $\sigma $ is the electric charge density of the plane sheet and ${{\varepsilon }_{0}}$ is the specific inductive capacity or the relative permittivity of the thin plane sheet. And then find the relation between the terms.
Complete step-by-step answer:
First of all let us take a look at what actually the electric field intensity means. The electric field Intensity at a point is given as the force felt by a unit positive charge kept at a particular point. Electric Field Intensity is basically a vector quantity. It is denoted by $E$. It can be written in a formula like,
$E=\dfrac{F}{q}$
For an infinite sheet of charge, the electric field is found to be perpendicular to the surface. Because of this, only the ends of a cylindrical Gaussian surface will be contributing to the electric flux.
Here in this question, it is mentioned that the sheet is plane uniformly charged. Therefore the electric field is given by the equation,
$E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$
By looking at this equation we are able to find that the electric field intensity is not having any sort of connection with the distance of the point from the plain sheet. Therefore the correct answer is option A.
So, the correct answer is “Option A”.
Note: Surface charge density denoted as $\sigma $, calculated in coulombs per square meter, $C{{m}^{-2}}$ is an amount of charge per unit area at any point on a surface charge distribution, in the case of a two dimensional surface. Since electric charge will be either positive or negative, charge density can be expressed as either positive or negative values.
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