Question

State Gauss theorem and apply it to find the electric field at a point due to (a) a line of charge (b) A plane sheet of charge (c) A Charged spherical conducting shell

Answer 1

Answer:

Electric field at a point due to$\oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {in }}}{\varepsilon_0}$

Explanation:

Step 1: Gauss's Law says that the charge enclosed divided by the permittivity determines the total electric flux out of a closed surface. The electric flux in a region is calculated by multiplying the electric field by the surface area that is projected in a direction perpendicular to the field.

Let qenc represent the total charge contained within the region of space inside the radius-r Gaussian spherical surface that is located at a distance r from the center. The relation for Gauss's rule is thus as follows: $4πr2E=qencϵ0.$

Step 2: GAUSS'S LAW: "Electric flow through an open surface depends only on the charges that are contained within it, net charges that are contained within Prematurity.

$Let we essume a slyndrical Qenclosed =d \cdot \ell \oint \vec{E} \cdot d \vec{A}=\int_{\text {cuered }} \vec{E} \cdot d \vec{A}+2 \int_{\text {plane }} \vec{E} \cdot d \vec{A} d \vec{D} E E$

$\oint_{E \cdot d A}=\int E d A=E \cdot 2 \pi r l=\frac{\text { Qènclosed }}{\varepsilon_0}=\frac{\lambda \cdot l}{\varepsilon_0}$

$\text { So. } E=\frac{\lambda}{2 \pi \varepsilon_0 \gamma}=\frac{2 k \lambda}{\gamma}$

$\lambda= linear charge density A=\text { Area }=(2 \pi r l)$

Therefor  electric field at a point due to $\oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {in }}}{\varepsilon_0}$

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