Electric potential — voltage and work

What just happened?

You moved the test charge around a circle $r = \text{const}$ around the source. Every point on this circle has the same potential, so the potential difference is zero and so is the work — regardless of the path length.

Potential of a point charge:

— the potential is directly proportional to the source charge and inversely proportional to distance. $Q$ — source charge (C), $r$ — distance (m), $k \approx 9\cdot10^{9}\ \tfrac{\mathrm{N\,m^2}}{\mathrm{C^2}}$.

Voltage and work:

Voltage is the potential difference between two points, and the work of moving a charge $q$ depends only on this difference — not on the shape of the path. When $V_1 = V_2$, then $W = 0$.

What does it mean in practice?

Equipotential surfaces are always perpendicular to the field lines. Moving along them, the field does no work — just as driving along a contour line in the mountains does not change your altitude. You do work only when crossing between different potential "contour lines".

Why you need this for the exam:

$V = kQ/r$ plus $W = q\,\Delta V$ solve most potential and voltage problems. Trick: compute the potential at each point, subtract — you get the voltage, and multiplying by the charge gives the work. Equipotential ⇒ $\Delta V = 0$ ⇒ $W = 0$.

Remember: the potential drops with distance — V = kQ/r, and the work depends only on the potential difference — W = q·ΔV. Along an equipotential the work is always zero.