Question

Two point charges qa and qb are located on the x-axis at x = a and x = b. FIGURE EX25.34 is a graph of V, the electric potential.

a. What are the signs of qa and qb?

b. What is the ratio ∙ qa/qb ∙?

c. Draw a graph of Ex, the x-component of the electric field, as a function of x

Short answer

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

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

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

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.

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,

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

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

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

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 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 expression for the electric filed for the point along the x-axis is,$E=\frac{2\eta L}{4\pi {\epsilon }_{0}x}$