Question

($\text{a}$) State Gauss’s law on electrostatics and derive an expression for the electric field due to a long straight thin uniformly charged wire (linear charge density $\lambda$ at a point lying at a distance $r$ from the wire.

($\text{b}$) The magnitude of electric field (in ${\text{NC}}^{–1}$) in a region varies with the distance $r$ (in $\text{m}$) as $E=10r+5$

By how much does the electric potential increase in moving from a point at $r=1\text{m}$ to a point at $r=10\text{m}$.

Answer 1

(a) Gauss’ law states that the electric flux of the resultant electric field through a closed surface is $\frac{1}{{\epsilon }_0}$ times of net charge enclosed by the surface

$\int \overrightarrow{E}.\overrightarrow{ds}=\frac{q_{net}}{{\epsilon }_0}$

where, $q$ = net charge enclosed by surface

${\epsilon }_0$ = permittivity of free space

Electric field due to a long straight wire

Consider an infinitely long thin wire with uniform linear charge density $\lambda$. To calculate the field imagine a cylindrical Gaussian surface as shown in the figure. Since the field is radial everywhere, therefore, flux through the two flat ends of the cylindrical Gaussian surface is zero.



Hence,

Flux through the Gaussian surface = Flux through the curved cylindrical part of the surface
$=E.2\pi rl$

The surface includes a charge equal to $\lambda l$, Gauss’s law then gives

$E\times 2\pi rl=\frac{\lambda l}{{\epsilon }_0}$

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$

$\text{Vectorially}\overrightarrow{E}=\frac{\lambda }{2\pi {\epsilon }_{0}r}\hat{n}$

$\hat{n}$ is the radial unit vector, in the plane normal to the wire passing through the point.

($\text{b}$) $E=10r+5$

The potential difference between the two points

$V_B-V_A=-{\int }_A^B\overrightarrow{E}.d\overrightarrow{r}$

$=-{\int }_{r=1}^{r=10}(10r+5).dr$

$=-{(\frac{10r^2}{2}+5r)}_1^{10}$

$=-\{[\frac{10\times 100}{2}+5\times 10]-[\frac{10}{2}+5]\}$

$=-(550-10)$

$=-540\text{volt}$

Hence potential increases by $–540\text{V}$ in moving from point $r=1\text{m}$ to $r=10\text{m}.$