91Ó°ÊÓ

Figure 24-31 shows a thin, uniform charged rod and three points at the same distance.

Considering radius values of the given point at different positions you can get the relation of the potentials as the radius is inversely proportional to the electric potential.

Formula:

The electric potential at a point due to a charged rod,

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r}$ ….. (i)

Where, q is the charge of the rod, r is the distance of the point from the rod, and ${\mathrm{Î\mu }}_0$is the permittivity of free space.

From equation (i), you can get that the potential at any point is inversely related to the distance value.

Let’s assume the small element of the rod of width dx at a distance x from the origin with charge dp. Therefore, the electric potential at pointa is given as,

$d{V}_a=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dp}{r_a}$ ….. (ii)

The distance r_a is the distance to the point which is written as,

$r_a=\sqrt{\left(x-{\left(L/2\right)}^2\right)+d^2}$ ….. (iii)

Substitute the value from equation (iii) in equation (ii).

$d{V}_a=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dp}{\sqrt{\left(x-{\left(L/2\right)}^2\right)+d^2}}$

Integrate this equation between x = 0 to x = d to get the value of V_a

role="math" localid="1661927777065" width="332" height="102">Va=λ4πε0∫0Ldxx-L/22+d2=λ4πε0Ind2x-L2+4d2+d2x-L0L

Therefore, the electric potential at point a is,

$\begin{array}{rcl}{V}_a& =& \frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[In\left(\frac{d\sqrt{L^2+4d^2}+d\left(L\right)}{d\sqrt{L^2+4d^2}-d\left(L\right)}\right)\right]\end{array}$ ….. (iv)

Now,the electric potential at point b is given as,

$d{V}_b=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{r_b}$ ….. (v)

The distance r_b is the distance to the point b which is written as,

$r_b=\sqrt{{\left(L-x\right)}^2+d^2}$ ….. (vi)

Substitute the value from equation (v) in the equation (vi) and writing charge in terms of charge density and width, we get

$d{V}_b=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{λ}dx}{\sqrt{{\left(L-x\right)}^2+d^2}}$

Integrate this equation between x = 0 to x = d to get the value of V_b .

role="math" localid="1661928338938" $\begin{array}{rcl}{V}_a& =& \frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^L\frac{dx}{\sqrt{{\left(L-x\right)}^2+d^2}}\\ & =& \frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[In\left(d\sqrt{{\left(x-L\right)}^2+4d^2}+d\left(x-L\right)\right)\right]}_0^L\end{array}$

Therefore, the electric potential at point b is,

role="math" localid="1661928580432" $V_b=\frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[In\left(\frac{2d^2}{d\sqrt{L^2+4d^2-d\left(L\right)}}\right)\right]$ ….. (vii)

Now,the electric potential at point c is given as,

$d{V}_c=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{r_c}$ ….. (viii)

The distance r_c is the distance to the point c which is written as,

$r_c=L-x+d$ ….. (ix)

Substitute the value from equation (ix) in the equation (viii) and write the charge in terms of charge density and width, you get

$d{V}_c=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{λ}dx}{L-x+d}$

Integrate this equation between x = 0 to x = d to get the value of V_c.

role="math" localid="1661928637951" $\begin{array}{rcl}{V}_c& =& \frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^L\frac{dx}{\sqrt{L-x+d}}\\ & =& \frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[-In\left(x-L+d\right)\right]}_0^L\end{array}$

Therefore, the electric potential at point c is,

$V_c=\frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left[In\left(\frac{L+d}{d}\right)\right]$ ….. (x)

Hence, the value of d is same for all the points a,b,and c. Therefore, from equation (iv), (vii) and (x), you can conclude that V_a > V_b > V_c .