
The $\sigma $ of a thin conducting sheet is $18 \times {10^{ - 8}}\,C{m^{ - 2}}.$ Find the electric field at a point near the sheet.
Answer
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Hint:In electrostatic, electric field is the field around a charge where the electric force can be experienced due to another charge, conductivity is the ability of a material to store the amount of electric charge on it, if charge is spread on a surface its referred as surface charge density and if its spread on volume its known as volume charge density. We will use the general formula of electric field due to a plane sheet at a distant point to calculate electric field near the sheet.
Formula used:
Conductivity of a plane sheet is defined as the charge spread across the whole sheet per unit area and it’s denoted as $\sigma $ or mathematically it can be expressed as
$\sigma = \dfrac{Q}{A}$
Where $Q$ is the total charge on the plane sheet and $A$ is the area of the plane sheet.
Electric field due to a plane sheet having negligible thickness at any point in space is independent of the distance of the point and the sheet and it’s calculated as $E = \dfrac{\sigma }{{2{ \in _0}}}$.
Complete step by step answer:
According to the question we have given that,
$\sigma = 18 \times {10^{ - 8}}C{m^{ - 2}}$
$\Rightarrow { \in _0} = 8.854 \times {10^{ - 12}}{C^2}{m^{ - 2}}{N^{ - 1}}$
On putting these values in formula, $E = \dfrac{\sigma }{{2{ \in _0}}}$ we get,
$E = \dfrac{{18 \times {{10}^{ - 8}}}}{{2 \times 8.854 \times {{10}^{ - 12}}}}$
$\Rightarrow E = \dfrac{{18 \times {{10}^4}}}{{17.708}}$
$\therefore E = 1.01 \times {10^4}N{C^{ - 1}}$
Hence, the electric field due to the plane sheet is $E = 1.01 \times {10^4}\,N{C^{ - 1}}$.
Note: It should be remembered that, electric field is a vector quantity which needs a direction to define it completely and, a sheet having positive conductivity shows the direction of electric field lines away from the sheet and a sheet having negative conductivity shows the direction of electric field lines is towards the plane sheet. Hence. In the given question the conductivity of the plane sheet was positive hence the direction of electric field lines will be away from the plane sheet.
Formula used:
Conductivity of a plane sheet is defined as the charge spread across the whole sheet per unit area and it’s denoted as $\sigma $ or mathematically it can be expressed as
$\sigma = \dfrac{Q}{A}$
Where $Q$ is the total charge on the plane sheet and $A$ is the area of the plane sheet.
Electric field due to a plane sheet having negligible thickness at any point in space is independent of the distance of the point and the sheet and it’s calculated as $E = \dfrac{\sigma }{{2{ \in _0}}}$.
Complete step by step answer:
According to the question we have given that,
$\sigma = 18 \times {10^{ - 8}}C{m^{ - 2}}$
$\Rightarrow { \in _0} = 8.854 \times {10^{ - 12}}{C^2}{m^{ - 2}}{N^{ - 1}}$
On putting these values in formula, $E = \dfrac{\sigma }{{2{ \in _0}}}$ we get,
$E = \dfrac{{18 \times {{10}^{ - 8}}}}{{2 \times 8.854 \times {{10}^{ - 12}}}}$
$\Rightarrow E = \dfrac{{18 \times {{10}^4}}}{{17.708}}$
$\therefore E = 1.01 \times {10^4}N{C^{ - 1}}$
Hence, the electric field due to the plane sheet is $E = 1.01 \times {10^4}\,N{C^{ - 1}}$.
Note: It should be remembered that, electric field is a vector quantity which needs a direction to define it completely and, a sheet having positive conductivity shows the direction of electric field lines away from the sheet and a sheet having negative conductivity shows the direction of electric field lines is towards the plane sheet. Hence. In the given question the conductivity of the plane sheet was positive hence the direction of electric field lines will be away from the plane sheet.
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