Solution:

Ans. Statement of Gauss Law 1

Derivation of electric field due to infinitely long straight uniformly charged wire

The surface integral of electric field over a closed surface is equal to $\frac{1}{\varepsilon_{0}}$ times the charge enclosed by the surface.

Alternatively,

$$ \begin{equation*} \oint \vec{E} \cdot \vec{d} s=\frac{q}{\varepsilon_{0}} \tag{1} \end{equation*} $$

Flux through the Gaussian surface

$=$ flux through the curved cylindrical part of the surface

$=\mathrm{E} \times 2 \pi r l$

$$ 1 / 2 $$

Charge enclosed by the surface $=\frac{\lambda}{2 \pi \varepsilon_{0} r}$

$$ =E \times(2 \pi r l) $$

$$ \begin{align*} & =\frac{\lambda l}{\varepsilon_{0}} \ \mathrm{E} & =\frac{\lambda}{2 \pi \varepsilon_{0} r} \end{align*} $$

[CBSE Marking Scheme, 2017]

Detailed Answer :

Gauss Law states that total electric flux over the closed surface $S$ will be $1 / \varepsilon_{0}$ times the total charge which is $\phi_{E}=\oint E d s=\frac{q}{\varepsilon_{0}}$

In a long straight wire with uniform charge per unit length $\lambda$, there should be electric field generated by charge distribution for cylindrical symmetry. Also, field to point will radially be away from the wire.

In this, cylindrical gaussian surface is co-axial with the wire of radius $R$ and length where symmetry implies to electric field generated by wire that will be perpendicular to curred-surface of cylinder, so as per Gauss’ law,

$$ E(R) \times 2 \pi R l=\frac{\lambda l}{\varepsilon_{0}} $$

where, $E(R)$ is electric field strength which acts at perpendicular distance $R$ from the wire.

In figure, left part shows electric flux through Gaussian surface while right part shows total charge enclosed by cylinder which is divided by $\varepsilon_{0}$.

Further, $\quad E(R)=\frac{\lambda}{2 \pi \varepsilon_{0} R}$

Here, the field points radially away from the wire when $\lambda>0$, and radially towards the wire when $\lambda<0$.

Commonly Made Error

• Some candidates do not know the correct expression. Few candidates are not sure about $\frac{1}{R}$ or $\frac{1}{R^{2}}$ (dependence of $E$ on $R$ ). AT Q. 4. A charge $Q$ is distributed uniformly over a metallic sphere of radius $R$. Obtain the expression for the electric field $(E)$ and electric potential $(V)$ at a point $0<x<R$.

Show or plot the variation of $E$ and $V$ with $x$ for $0<x<2 R$. A&E [Foreign I, II, III 2017]

Ans. Expression for electric field

$1 \frac{1}{2}$

Expression for potential

Plot of graph ( $E$ Vs $r$ ) $1 / 2$

Plot of graph $(V V s r)$

By Gauss law

$1 / 2$

$q=0$ in interval $0<x<\mathrm{R}$

$$ E=0 $$

or

$E=-\frac{d V}{d r}$

Hence, $V=$ constant $=V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}$

[Even if a student draws $E$ and $V$ for $0<r<R$ award $1 / 2+1 / 2$ mark]

[CBSE Marking Scheme, 2017]