Electrostatic potential
Work against the electric field
Lift a stone and you store gravitational potential energy. Push two like charges closer together and you do the same thing: you store electrostatic potential energy. Let us see how.
A charge Q is fixed at the origin. We carry a small test charge q, slowly, from a point R to a point P closer to Q. Q pushes it away, so we must push it in.
Our external force is exactly equal and opposite to the electric force, so q never speeds up. The work done by our force is stored as potential energy.
So the work done by the external force from R to P equals the rise in potential energy, U_P − U_R. It is also equal to minus the work done by the electric force.
The Coulomb force is conservative, like gravity. Carry q along a straight path or a winding one: the work is the same. It depends only on the end points R and P.
Only differences in potential energy matter, so we may choose where U is zero. The usual choice is infinity, where the charges are too far apart to feel each other.
Then the potential energy of q at P is the work done by an external force in bringing q from infinity to P.
This work is proportional to q. Double the test charge and the work doubles. So divide by q, and we get something that depends only on Q and the point: the electrostatic potential V.
The potential at a point is the work done by an external force in bringing a unit positive charge, without acceleration, from infinity to that point.
Its unit is the joule per coulomb, called the volt. Between any two points, the work needed is W = q(V_P − V_R). If 3 joules carry 1.5 coulombs from R to P, then V_P − V_R is 2 volts.
Potential due to a point charge
Now let us find the potential at a distance r from a single point charge Q. Put Q at the origin, and bring a unit positive charge in from infinity along a straight line.
At a distance r′, the force on the unit charge is Q/4πε₀r′², pointing outward. To move it in by a small step, we must push against this force.
Adding up the work for every small step, from infinity down to r, gives Q/4πε₀r. So V(r) = Q/4πε₀r, or simply kQ/r.
The work does not depend on the path, so this holds in every direction. V depends only on the distance r: every point on a sphere around Q has the same potential.
For a positive Q, V is positive. For a negative Q, V is negative: the field pulls the unit charge in, so the external agent does negative work.
Compare the two graphs. The field falls off as 1/r², but the potential falls more slowly, as 1/r. Double the distance: V halves, but E becomes a quarter.
Worked example: a charge of 4 × 10⁻⁷ C. The potential 9 cm away is 9 × 10⁹ × 4 × 10⁻⁷ divided by 0.09. That is 4 × 10⁴ volts.
How much work does it take to bring a charge of 2 × 10⁻⁹ C from infinity to that point? W = qV = 2 × 10⁻⁹ × 4 × 10⁴, which is 8 × 10⁻⁵ joules, along any path.
Remember: potential is a scalar. It has a size and a sign, but no direction. That makes it much easier to add up than the field, as we will see next.
Try it: carry a test charge
- Drag the green test charge q₀ away from the charge. At twice the distance, V halves; compare it with the kq/r readout.
- Press “Set A at q₀”, then wander about. The work summed along your path always equals q₀(V − V_A).
- Carry q₀ round a closed loop back to A: the total work comes back to zero.
- Make the charge negative with the slider: the potential becomes negative everywhere.
Twice as far
At 3 cm from a +2 nC charge, V = 600 V and E = 20 kV/m. What are V and E at 6 cm?
- V = 300 V, E = 5 kV/m
- V = 150 V, E = 5 kV/m
- V = 300 V, E = 10 kV/m
- V = 150 V, E = 10 kV/m
Answer
A. V = kQ/r falls as 1/r: doubling r halves it, 600 V → 300 V. E = kQ/r² falls as 1/r²: doubling r quarters it, 20 → 5 kV/m. Check: 9 × 10⁹ × 2 × 10⁻⁹ / 0.06 = 300 V.
Does the path matter?
A 2 μC charge is carried from point A, at 100 V, to point B, at 400 V. How much work must an external agent do, (i) along a straight line and (ii) along a long winding path?
- 6 × 10⁻⁴ J along both paths
- 8 × 10⁻⁴ J along both paths
- 6 × 10⁻⁴ J along the straight line, more along the winding path
- 1 × 10⁻³ J along both paths
Answer
A. W = q(V_B − V_A) = 2 × 10⁻⁶ × (400 − 100) = 6 × 10⁻⁴ J. The electrostatic force is conservative, so the work depends only on the end points, not on the path. (8 × 10⁻⁴ J uses V_B alone; 1 × 10⁻³ J adds the potentials.)
Summary
- Potential (from infinity):
V = W/q - Work between two points:
W(R→P) = q(V_P − V_R) - Point charge:
V = Q/4πε₀r = kQ/r - Unit:
1 V = 1 J/C
- The electrostatic force is conservative: the work done against it depends only on the end points, so a potential energy can be defined.
- The potential at a point is the work done by an external force in bringing a unit positive charge, without acceleration, from infinity to that point.
- For a point charge, V ∝ 1/r while E ∝ 1/r². V is positive near a positive charge and negative near a negative one.
- Potential is a scalar: it has a sign but no direction.