(a) Use Gauss’ law to derive the expression for the electric field \left ( \vec{E} \right )due to a straight uniformly charged infinite line of charge density \lambda \; C/m\cdot
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r_{1}\, to\, r_{2}\left ( r_{2}> r_{1} \right )\cdot





Answers (1)
S safeer

a) Electric field due to an infinitely long straight charged wire

Consider infinitely long straight charged wire of linear charge density \lambda \, C/m
For calculating electric field consider an imaginary cylindrical Gaussian Surface of radius r and length l.
Here the field is radial everywhere, so flux through the two ends of cylinder is Zero.
At the Gaussian cylindrical surface, the electric field E is normal to the surface at every point. The magnitude of E depends only on radius 'r', so it is constant.
Therefore flux through Gaussian surface
              = E\times 2\pi rl---(1)
According to Gauss's law , flux

\phi = \frac{charge\; enclosed}{\varepsilon _{0}}
Here, total charge enclosed = linear charge density \times length
                                        = \lambda l
Therefore flux \phi= \frac{\lambda l}{\varepsilon _{0}}---(2)
Using equation (1) and (2)  E\times 2\pi rl= \frac{\lambda l}{\varepsilon _{0}}
That is ,

E= \frac{\lambda }{2\pi \varepsilon _{0}r}
The vector notation is

\vec{E}= \frac{\lambda }{2\pi \varepsilon _{0}r}\hat{n}
where \hat{n} is the radial unit vector normal to the line charge

E\alpha \frac{1}{r}
c) work done to move the charge 'q' through a small displacement 'dr'
dW= F\cdot dr
We know , F= qE
where E = electric field
Therefore dW= qEdr\cos \theta = qEdr\left ( \theta = 0^{\circ} \right )
Substitute the value of E
    dW= q\times \frac{\lambda }{2\pi \varepsilon _{0}r}dr
work done in moving the charge from r1 to r2 \left ( r_{2}> r_{1} \right ) is
W=\int_{r_{1}}^{r_{2}}dW= \int_{r_{1}}^{r_{2}}q\, \frac{\lambda }{2\pi \varepsilon _{0}r}dr
W=q\, \frac{\lambda }{2\pi \varepsilon _{0}}\left [ l_{n} r_{2}- l_{n} r_{1}\right ]
W=q\, \frac{\lambda }{2\pi \varepsilon _{0}}ln\frac{r_2}{r_1}

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