Question

The long rod shown in Figure 16.76 has length L and carries a uniform charge −Q. Calculate the potential difference ${\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{C}}$. All of the distances are small compared to L. Explain your work carefully

Short answer

The potential difference is positive whose value is $\frac{2\mathrm{KQ}}{\mathrm{L}}\ln \left(\frac{\mathrm{c}}{\mathrm{b}}\right)$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The length of the rod is, L .
• The charge on the rod is, -Q.

Concept/Significance of the potential difference.

The potential difference is the amount of effort required to move a charge from one location in an electric field to another.

It is the amount of effort or energy necessary to transport an electron in a circuit or across a geographic region of an electrical field.

Determination of the potential difference VA-VC

The potential difference between point A and C.

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{C}}=({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}})-({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}})\dots \left(\mathrm{i}\right)$

The potential difference between point A and B is given by,

${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=0$

The potential difference between point B and C is given by,

localid="1657097638094" ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}}={\int }_{\mathrm{xB}}^{\mathrm{xC}}{\mathrm{E}}_{\mathrm{x}}\mathrm{dx}.........\left(\mathrm{ii}\right)$

Here, $E_x$ is the electric field in x-direction.

The electric field near the center of rod is given by,

${\mathrm{E}}_{\mathrm{x}}=\frac{-2\mathrm{KQ}}{\mathrm{xL}}$

Here, K is the coulomb constant, Qis the charge on the rod,x is the distance betweentwo point and L is the length of the rod.

Substitute all the values in the equation (ii),

$$\begin{gathered} {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}}=\underset{{\mathrm{x}}_{\mathrm{B}}}{\overset{{\mathrm{x}}_{\mathrm{C}}}{\int }}\frac{{\displaystyle \frac{2\mathrm{KQ}}{\mathrm{xL}}}}{}\mathrm{dx} \\ =-\frac{2\mathrm{KQ}}{\mathrm{xL}}\underset{\mathrm{b}}{\overset{\mathrm{c}}{\int }}\frac{1}{\mathrm{x}}\mathrm{dx} \\ =\frac{2\mathrm{KQ}}{\mathrm{xL}}\mathrm{In}{\left(\mathrm{x}\right)}_{\mathrm{b}}^{\mathrm{c}} \\ =\frac{2\mathrm{KQ}}{\mathrm{xL}}\mathrm{In}\left(\frac{\mathrm{c}}{\mathrm{b}}\right) \end{gathered}$$

The value oflocalid="1660977943817" $\mathrm{c}/\mathrm{b}<1$so its value is negative.

Thus, the potential difference is positive whose value is $\frac{2\mathrm{KQ}}{\mathrm{L}}\ln \left(\frac{\mathrm{c}}{\mathrm{b}}\right)$.