Q.The electrostatic potential is given as a function of $x$ in figure (a), where $V$ rises linearly with $x$ across four labelled regions A, B, C, D between $x=0$ and $x=0.6\ \text{m}$, and in figure (b), a similar $V$ versus $x(\text{cm})$ graph over $x=1$ to $5\ \text{cm}$. Calculate the corresponding electric fields in regions A, B, C and D for figure (a), and plot the electric field as a function of $x$ for figure (b).

A circle drawn around a point charge is an equipotential surface, so moving a charge once around it and back to the start involves zero net work.

A point charge +q at the centre produces a potential $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ that depends only on the distance r from the charge. Every point on the circumference of a circle of radius r is at the same distance r from the centre, so every point on the circle is at the same potential — the circle is an equipotential curve.

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Work done in moving a charge between two points depends only on the potential difference between them; since A and B are equidistant from P, they are at the same potential, so W = 0.

The electric potential due to a point charge $+q$ at P, at a distance $r$ from it, is

$V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Since A and B both lie on the circumference of a circle centred at P, they are at the SAME distance $r$ from P, so

$V_A = V_B$

The work done in moving a charge $q$ from A to B is

$W = q(V_B - V_A) = q(0) = 0$

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