Electrostatic Potential and CapacitanceElectrostatic Potential
The work needed to bring a charge to a point depends on the charge itself. At any point the force on a test charge is , so twice the charge needs twice the work. That makes the work a property of the test charge as well as of the point. Can we strip the test charge out and keep a number that belongs to the point alone?
There is a point P near a charged dome at the school science exhibition. Bring different test charges from infinity to P, one at a time. Each one plots its work W against its charge q.
The amber points lie on the purple line through the origin. Its slope, W ÷ q, is the same 30 µJ per nC for every test charge — the work done per unit charge. That ratio is a property of the point P alone.
Work per Unit Charge
Dividing the work by the charge gives a new quantity, the electrostatic potential V. From Eq. (2.1) and (2.2):
Again only differences matter, and again we choose the potential to be zero at infinity. Then:
Electrostatic potential. The electrostatic potential V at a point is the work done by an external force in bringing a unit positive charge, without acceleration, from infinity to that point.
The unit is the volt (V), named after the Italian physicist Alessandro Volta (1745–1827), who built the first battery — a tall pile of metal discs with moist cardboard between them.
Back at the dome: W ÷ q was 30 µJ per nC. Since 1 µJ per nC is 1000 J/C, the potential at P is
And bringing a 5 nC charge from infinity to P needs
Note
Strictly, the test charge should be vanishingly small, , so that it cannot disturb the charges that make the field; the potential is the ratio . And the external force must at every point be equal and opposite to the electric force on the test charge. Fig. 2.2 of your textbook makes the same point you found with the paths: the work done by the field depends only on the start and the end.