Potential due to a Point Charge

The derivation CBSE sets

Take a point charge $Q$ at the origin and a point $P$ at distance $r$. Bring a unit positive test charge from infinity to $P$ along the radial line, which is legitimate because the work is path independent.

At an intermediate point $P'$ at distance $r'$, the electrostatic force on a unit positive charge is

$$\vec F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r'^{2}}\hat r'$$

Work done by the external agent for a displacement $\Delta r'$ is $\Delta W=-\dfrac{Q}{4\pi\varepsilon_0 r'^{2}}\Delta r'$; the minus sign appears because the displacement is inward ($\Delta r'<0$) while the work is positive. Integrating from $r'=\infty$ to $r'=r$,

$$W=-\int_{\infty}^{r}\dfrac{Q}{4\pi\varepsilon_0 r'^{2}}\,dr'=\left[\dfrac{Q}{4\pi\varepsilon_0 r'}\right]_{\infty}^{r}=\dfrac{Q}{4\pi\varepsilon_0 r}$$

By definition this work is the potential at $P$:

$$V(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$$

Reading the result

• The formula holds for either sign of $Q$. For $Q<0$ it gives $V<0$: the external work per unit positive charge is negative, because the electrostatic force and the displacement from infinity point the same way.
• It is consistent with the choice $V_\infty=0$: as $r\to\infty$, $V\to0$.
• Any arbitrary infinitesimal displacement resolves into a radial part and a perpendicular part; the perpendicular part does no work, which is why any path gives the same answer.

Graphs

• $V$ against $r$: a hyperbola, $V\propto1/r$, positive branch for $Q>0$, falling steeply near the charge and approaching zero asymptotically.
• $E$ against $r$: also decreasing, but as $1/r^{2}$, so it falls faster and lies below the potential curve at large $r$ (in the natural units $Q/4\pi\varepsilon_0$ per metre and per square metre used in NCERT Fig. 2.4).
• $V$ against $1/r$: a straight line through the origin with slope $\dfrac{Q}{4\pi\varepsilon_0}$. CBSE has used this: if two such lines make angles $60^\circ$ and $30^\circ$ with the $1/r$ axis, then $\dfrac{Q_1}{Q_2}=\dfrac{\tan60^\circ}{\tan30^\circ}=\dfrac{\sqrt3}{1/\sqrt3}=3$, i.e. $Q_1:Q_2=3:1$.

Sign of a potential difference

For $+Q$ at $O$ with $OA=r_A<OB=r_B$:

$$V_A-V_B=\dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{r_A}-\dfrac{1}{r_B}\right)>0$$

so $V_A-V_B$ is positive — potential decreases as you move away from a positive charge. State the reason, not just the sign.

Diagram to describe in words

A positive charge $Q$ at the origin, with a straight radial line running out to $P$ at distance $r$. Mark an intermediate point $P'$ at $r'$ with the outward repulsive force on the unit test charge and the small inward displacement $\Delta r'$. Beside it, a graph with $r$ on the $x$-axis carrying two curves, the shallower $1/r$ potential curve and the steeper $1/r^{2}$ field curve, both labelled.

Worked pattern from the textbook

For $Q=4\times10^{-7}\ \text{C}$ at $9\ \text{cm}$: $V=\dfrac{9\times10^{9}\times4\times10^{-7}}{0.09}=4\times10^{4}\ \text{V}$. Bringing $q=2\times10^{-9}\ \text{C}$ from infinity to that point needs $W=qV=8\times10^{-5}\ \text{J}$, and the answer does not depend on the path.

Standard results