Q.A point charge $Q$ sets up a radial electric field. A small test charge $q$ is carried around two different closed paths in this field. The first closed path is made of segments that run either exactly along the field lines (radially outward or inward) or exactly perpendicular to them (along circular arcs), returning to its starting point. The second closed path is a rectangular loop enclosing the same area as the first path. How does the work done by the electric force on the test charge over the first closed path compare with that over the second closed path?

Concept

An electrostatic field is conservative: $\oint \vec{E}\cdot d\vec{l} = 0$. The work done by the field on a charge $q$ over a closed loop is $W = q\oint \vec{E}\cdot d\vec{l} = 0$, whatever the loop's shape or the area it encloses.

Steps

1. First path: on the radial segments the force is along (or opposite to) the motion so work is done, while on the arc segments the force is perpendicular to the motion so the work is zero. Summed around the full closed path these contributions cancel to zero. …

Method: Work Done by the Electrostatic Force Over a Closed Path

Use this method for ANY question that asks you to compare or compute the work done by the electric force as a charge is carried around one or more closed loops — regardless of the loop's shape.

Steps

Step 1: Recognise the defining property of the electrostatic field

The electrostatic field is conservative:

for ANY closed path, in ANY electrostatic field (uniform, radial, or otherwise) — this is a structural property of every static Coulomb field, not something you derive per-problem.

Step 2: Translate into work done

Since work done by the field on a charge $q$ is $W = q\displaystyle\oint \vec{E}\cdot d\vec{l}$, this immediately gives

no matter how the loop is shaped, how large an area it encloses, or how many straight/curved segments it has.

Step 3: Do not be misled by segment-by-segment work …