State Gauss's Law in electrostatics. Using Gauss's Law, derive the expression for the electric field intensity due to an infinitely long straight uniformly charged wire.

Q: State Gauss's Law in electrostatics. Using Gauss's Law, derive the expression for the electric field intensity due to an infinitely long straight uniformly charged wire.

Gauss's Law:\nGauss's law states that the total electric flux $(\PhiE)$ passing through any closed surface in vacuum is equal to $\frac{1}{\varepsilon0}$ times the net charge $(q)$ enclosed by that surface. $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon0}$$ Derivation for an Infinitely Long Straight Wire: Let an infinitely long straight wire have a uniform linear charge density $\lambda$. Consider a cylindrical Gaussian surface of radius $r$ and length $l$ coaxial with the wire. The cylindrical Gaussian surface has three parts: two circular flat cap surfaces ($S1, S2$) and one curved surface ($S3$). For the flat ends ($S1, S2$), $\vec{E}$ and $d\vec{A}$ are perpendicular ($\theta = 90^\circ$), so flux $\Phi1 = \Phi2 = 0$. For the curved surface ($S3$), $\vec{E}$ and $d\vec{A}$ are in the same direction ($\theta = 0^\circ$). Total Flux through the Gaussian Surface: $$\PhiE = \oint{S3} E \cdot dA \cdot \cos 0^\circ = E \cdot (2\pi r l)$$ By Gauss's Law: $$E \cdot (2\pi r l) = \frac{q}{\varepsilon0}$$\nSince enclosed charge $q = \lambda \cdot l$: $$E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon0}$$ $$E = \frac{\lambda}{2\pi \varepsilon0 r}$$\nIn vector form: $\vec{E} = \frac{\lambda}{2\pi \varepsilon0 r} \hat{r}$

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Accepted Answer

Gauss's Law:
Gauss's law states that the total electric flux $(\PhiE)$ passing through any closed surface in vacuum is equal to $\frac{1}{\varepsilon0}$ times the net charge $(q)$ enclosed by that surface. $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon0}$$ Derivation for an Infinitely Long Straight Wire: Let an infinitely long straight wire have a uniform linear charge density $\lambda$. Consider a cylindrical Gaussian surface of radius $r$ and length $l$ coaxial with the wire. The cylindrical Gaussian surface has three parts: two circular flat cap surfaces ($S1, S2$) and one curved surface ($S3$). For the flat ends ($S1, S2$), $\vec{E}$ and $d\vec{A}$ are perpendicular ($	heta = 90^\circ$), so flux $\Phi1 = \Phi2 = 0$. For the curved surface ($S3$), $\vec{E}$ and $d\vec{A}$ are in the same direction ($	heta = 0^\circ$). Total Flux through the Gaussian Surface: $$\PhiE = \oint{S3} E \cdot dA \cdot \cos 0^\circ = E \cdot (2\pi r l)$$ By Gauss's Law: $$E \cdot (2\pi r l) = \frac{q}{\varepsilon0}$$
Since enclosed charge $q = \lambda \cdot l$: $$E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon0}$$ $$E = \frac{\lambda}{2\pi \varepsilon0 r}$$
In vector form: $\vec{E} = \frac{\lambda}{2\pi \varepsilon0 r} \hat{r}$

NCERT · Class 12 · Physics · Electric Charges and FieldsState Gauss's Law in electrostatics. Using Gauss's Law, derive the expression for the electric field intensity due to an infinitely long straight uniformly charged wire.