Question

a. An infinitely long sheet of charge of width L lies in the x y plane between x=-L / 2 and x=L / 2. The surface charge density is $\eta .$ Derive an expression for the electric field $\overrightarrow{E}$ along the x-axis for points outside the sheet $(x>L/2).$

b. Verify that your expression has the expected behavior if$x\gg L$.

Hint: $\ln (1+u)\approx uifu\ll 1.$

c. Draw a graph of field strength E versus x for x>L / 2.

Short answer

expression for the electric filed $\overrightarrow{E}$along the x-axis for points outside the sheets is

$E=\frac{2\eta }{4\pi {\epsilon }_0}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)\text{.}$

Therefore, if$x>>>L$ then electric field is inversely proportional to the distance of the point from the sheet along the $x-axis$.

Therefore, graph of field strength E versus x is shown in figure I.

Step-by-step solution

part(a) step 1:given information

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

$d\lambda =\eta ds$

$\lambda$is the linear charge density of the strip.

$\eta$is the surface charge density of the sheet.

$dw$is the small width of the assume strip.

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,

$dE=\frac{2\Delta \lambda }{4\pi {\epsilon }_{0}r}$

Substitute $\eta$d s for d $\lambda$and (x-s) for r

$dE=\frac{2\eta ds}{4\pi {\epsilon }_0(x-s)}$

Substitute \eta d s for d \lambda and (x-s) for r.

$dE=\frac{2\eta ds}{4\pi {\epsilon }_0(x-s)}$

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

$\int E={\int }_{-L/2}^{+L/2}\frac{2\eta ds}{4\pi {\epsilon }_0(x-s)}$

$E=\frac{2\eta }{4\pi {\epsilon }_0}{\int }_{-L/2}^{L/2}\frac{ds}{(x-s)}$

$=\frac{2\eta }{4\pi {\epsilon }_0}(-1)\left[\ln \right(x-s){]}_{-L/2}^{L/2}$

$=\frac{2\eta }{4\pi {\epsilon }_0}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

part(b) step 1:given information

The width of an infinitely long sheet is L.

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

$E=\frac{2\eta }{4\pi {\epsilon }_0}\ln \left(\frac{x+\frac{L}{2}}{x-\frac{L}{2}}\right)$

$=\frac{2\eta }{4\pi {\epsilon }_0}\ln \left(\frac{1+\frac{L}{2x}}{1-\frac{L}{2x}}\right)$

$=\frac{2\eta }{4\pi {\epsilon }_0}\left[\ln \left(1+\frac{L}{2x}\right)-\ln \left(1-\frac{L}{2x}\right)\right]$

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

$E=\frac{2\eta }{4\pi {\epsilon }_0}\left[\frac{L}{2x}-\ln 1\right]$

$=\frac{2\eta L}{4\pi {\epsilon }_{0}x}$

part(c) step 1:given information

The width of an infinitely long sheet is L.

The expression for the electric filed for the point along the x-axis is,

$E=\frac{2\eta L}{4\pi {\epsilon }_{0}x}$