Question
State Gauss᾿s theorem in electrostatics. Using this theorem, derive an expression for the electric field due to an infinitely long straight wire of linear charge density $\lambda$.
✓
Answer
(I) Gauss’s theorem: The flux of electric field through any closed surface is $\frac{1}{\varepsilon_0}$ times the total charge enclosed by the closed surface.
$\phi=\frac{q}{\varepsilon_0}$ ..(1)
By definition, the total electric flux through the closed surface is given by
$\phi=\oint \vec{E} \cdot \overrightarrow{d s}$
$\therefore$ From (1) and (2), Gauss's theorem may be expressed as follows
$\phi=\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_0}$
The surface integral of electric field over a closed surface is equal to $\frac{1}{\varepsilon_0}$ times the total charge enclosed by the surface.
Application of Gauss’s theorem
To find electric field due to a line charge let us consider an infinitely long line charge placed along XX’ axis with linear charge density λ. Our aim is to find electric field intensity at a point P distant r from the line charge. We draw a cylindrical surface of radius r and length l coaxial with the line charge. The net flux through the cylindrical gaussian surface i.e.
$
\begin{aligned}
\phi= & \phi \vec{E} \cdot \overrightarrow{d s}=\int_{L C F} \vec{E} \cdot \overrightarrow{d s}+\int_{C s} \vec{E} \cdot \overrightarrow{d s}+\int_{R C F} \vec{E} \cdot \overrightarrow{d s} \\
& \quad=\int_{L C F} E d s \cos 90^{\circ}+\int_{C s} E d s \cos 0^{\circ}+\int_{R C F} E d s \cos 90^{\circ} \\
& \phi=\int_{C s} E d s \cos 0^{\circ}=E \cdot 2 \pi r l
\end{aligned}
$
The charge enclosed by the gaussian surface is q = N
Using Gauss’s theorem from equations (1) and (2)
$E(2 \pi r l)=\frac{\lambda l}{\varepsilon_0} \Rightarrow E=\frac{\lambda}{2 \pi \varepsilon_0 r}$
$\phi=\frac{q}{\varepsilon_0}$ ..(1)
By definition, the total electric flux through the closed surface is given by
$\phi=\oint \vec{E} \cdot \overrightarrow{d s}$
$\therefore$ From (1) and (2), Gauss's theorem may be expressed as follows
$\phi=\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_0}$
The surface integral of electric field over a closed surface is equal to $\frac{1}{\varepsilon_0}$ times the total charge enclosed by the surface.
Application of Gauss’s theorem
To find electric field due to a line charge let us consider an infinitely long line charge placed along XX’ axis with linear charge density λ. Our aim is to find electric field intensity at a point P distant r from the line charge. We draw a cylindrical surface of radius r and length l coaxial with the line charge. The net flux through the cylindrical gaussian surface i.e.
$
\begin{aligned}
\phi= & \phi \vec{E} \cdot \overrightarrow{d s}=\int_{L C F} \vec{E} \cdot \overrightarrow{d s}+\int_{C s} \vec{E} \cdot \overrightarrow{d s}+\int_{R C F} \vec{E} \cdot \overrightarrow{d s} \\
& \quad=\int_{L C F} E d s \cos 90^{\circ}+\int_{C s} E d s \cos 0^{\circ}+\int_{R C F} E d s \cos 90^{\circ} \\
& \phi=\int_{C s} E d s \cos 0^{\circ}=E \cdot 2 \pi r l
\end{aligned}
$
The charge enclosed by the gaussian surface is q = N
Using Gauss’s theorem from equations (1) and (2)
$E(2 \pi r l)=\frac{\lambda l}{\varepsilon_0} \Rightarrow E=\frac{\lambda}{2 \pi \varepsilon_0 r}$
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