Question

Figure 16.60 shows a portion of a long, negatively charged rod. You need to calculate the potential difference ${\mathit{V}}_{\mathbf{D}}\mathbf{-}{\mathit{V}}_{\mathbf{c}}$.

(a) What is the direction of the path $\left(+y\mathrm{or}-y\right)$?

(b) What is the sign of${\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}$?

Short answer

a). The potential difference is a property of $\mathrm{E}$generated by the source charges. It is independent of the sign or magnitude of the test charge.

b). ${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}$, The direction of the path is along the -y direction.

Step-by-step solution

Direction of the path.

The potential difference is a property of$\mathrm{E}$ generated by the source charges. It is independent of the sign or magnitude of the test charge. The sign of the potential difference${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}$ due to the rod does not depend on the sign of the test charge.

Sign of VA-VB.

Since to determine, ${\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}$, $\mathrm{A}$is the final point and B is the initial point, so the path is from $\mathrm{B}\mathrm{to}\mathrm{A}$.

Notice that with the source charge on the rod being negative, it generates a downward electric field. By inspection the vector $\mathrm{E}$is parallel to the path,

So,$\square \mathrm{V}={\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}<0$.

Hence, the direction of the path is along the -y direction.