91Ó°ÊÓ

The width of an infinitely long sheet is L.

Consider a strip of small width d w at a distance s from the center of the sheet. The linear charge distribution on that strip is,

$d\mathrm{λ}=\mathrm{η}ds$

$\mathrm{λ}$is the linear charge density of the strip.

$\mathrm{η}$is the surface charge density of the sheet.

d w is the small width of the assume strip.

Now, consider a point P on the axis outside the sheet at distance x from the center of the sheet. So, the distance of the point P from the strip is,

$r=x-s$

r is the distance of the point P from the strip.

Formula to calculate the electric field at point P due to the strip is,

$dE=\frac{2\mathrm{Δ}\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}$

Substitute $\mathrm{η}$d s for d $\mathrm{λ}and(x-s)$for r.

$dE=\frac{2\mathrm{η}ds}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x-s)}$

Integrate the above equation to find the net field strength due to the whole sheet at point P.

$∫E={∫}_{-L/2}^{+L/2}\frac{2\mathrm{η}ds}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0(x-s)}$

$E=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_{-L/2}^{L/2}\frac{ds}{(x-s)}$

$=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}(-1)\left[\ln \right(x-s){]}_{-L/2}^{L/2}$

$=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

The expression for the electric field at point on the x-axis outside the sheet is,

$E=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

$=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\ln \left(\frac{1+\frac{L}{2x}}{1-\frac{L}{2x}}\right)$

$=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\ln \left(1+\frac{L}{2x}\right)-\ln \left(1-\frac{L}{2x}\right)\right]$

If $x>>>L$, then $\frac{L}{2x}<<<1.$So, using In property that is, $\ln (1+u)≈uifu<<<1$

$E=\frac{2\mathrm{η}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[\frac{L}{2x}-\ln 1\right]$

$=\frac{2\mathrm{η}L}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}x}$

The expression for the electric filed for the point along the x-axis is,$E=\frac{2\mathrm{η}L}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}x}$