2.3 POTENTIAL DUE TO A POINT CHARGE

2.3

Work done in bringing a unit positive test charge from infinity to the point $P$, against the repulsive force of charge $Q(Q>0)$ , is the potential at $P$ due to the charge $Q$.

Consider a point charge $Q$ at the origin (Fig. 2.3). For definiteness, take $Q$ to be positive. We wish to determine the potential at any point $P$ with position vector $r$ from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point $P$. For $Q>0$, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point $P$.

At some intermediate point $P^{\prime }$ on the path, the electrostatic force on a unit positive charge is $$\frac{Q\times 1}{4\pi {\epsilon }_0{r}^{\prime 2}}{\hat{\mathbf{r}}}^{\prime }$$ (2.5)
where ${\hat{\mathbf{r}}}^{\prime }$ is the unit vector along $O{P}^{\prime }$.

Work done against this force from ${\mathbf{r}}^{\prime }$ to ${\mathbf{r}}^{\prime }+\mathrm{\Delta }{\mathbf{r}}^{\prime }$ is $$\mathrm{\Delta }W=-\frac{Q}{4\pi {\epsilon }_0{r}^{\prime 2}}\mathrm{\Delta }{r}^{\prime }$$ (2.6)

The negative sign appears because for $\mathrm{\Delta }{r}^{\prime }<0$, $\mathrm{\Delta }W$ is positive.

Total work done ($W$) by the external force is obtained by integrating Eq. (2.6) from $r^{\prime }=\mathrm{\infty }$ to $r^{\prime }=r$,

$$W=-{\int }_{\mathrm{\infty }}^r\frac{Q}{4\pi {\epsilon }_0{r}^{\prime 2}}d{r}^{\prime }={\frac{Q}{4\pi {\epsilon }_0{r}^{\prime }}|}_{\mathrm{\infty }}^r=\frac{Q}{4\pi {\epsilon }_{0}r}$$ (2.7)

This, by definition is the potential at $P$ due to the charge $Q$

$$V(r)=\frac{Q}{4\pi {\epsilon }_{0}r}$$ (2.8)

Equation (2.8) is true for any sign of the charge $Q$, though we considered $Q>0$ in its derivation. For $Q<0$, $V<0$, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point $P$ is positive. [This is as it should be, since for $Q<0$, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to $P$) are in the same direction.] Finally, we note that Eq. (2.8) is consistent with the choice that potential at infinity be zero.

2.4

Variation of potential $V$ with $r$ [in units of $(Q/4\pi {\epsilon }_0)m^{-1}$ ] (blue curve) and field with $r$ [in units of $(Q/4\pi {\epsilon }_0)m^{-2}$] (black curve) for a point charge $Q$.

Figure (2.4) shows how the electrostatic potential $(\propto 1/r)$ and the electrostatic field $(\propto 1/r^2)$ varies with $r$.