Electric Potential Due to a Point Charge

In electrostatics, electric potential helps us measure how much work is needed to bring a charge from infinity to a point in an electric field.
For a point charge, this potential depends only on the distance from the charge and not on direction, which makes the field spherically symmetric.

The work done by an external agent in bringing a unit positive test charge slowly from infinity to a point in an electric field, against the electrostatic force, is called the electric potential at that point.

\[V(r)=\frac{1}{4\pi\varepsilon_0K}\frac{q}{r}\]

• V(r) = electric potential at distance rr from the charge
• q = source charge
• ε₀ = permittivity of free space
• K = dielectric constant of medium
• Reference is taken such that V(∞) = 0.

• Consider a point charge +q at the origin O.
• We wish to find the potential at a point P at a distance r from O.
• Place a unit positive test charge at distance x from O along OP.

Electrostatic force on a unit positive charge at a distance x:

• Direction of F is radially outward (away from +q).

Small work done in moving the test charge:

• Move the unit positive charge from x to x + dx towards the charge.
• Displacement is towards O, force is away from O, so:
  dW = −F dx = −\[\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}dx\]
  (Negative sign appears because displacement is opposite to the direction of force.)

Total work done from infinity to distance r:

Electric potential at distance r:

For a unit test charge, V(r) = W, so

• Consider the same point charge +q embedded in a medium of dielectric constant K.
• Effective permittivity is ε = Kε₀.

Force on unit positive test charge at distance x:

Small work done:

Total work done:

Potential at distance r in the medium:

For air/vacuum, K ≈ 1, so it reduces to the earlier formula.

• If q > 0: V(r) > 0 (positive potential).
• If q < 0: V(r) < 0 (negative potential).
• At r → ∞: V → 0.
• Potential depends only on distance r, not on direction → spherically symmetric.
• Equipotential surfaces are concentric spheres around the point charge.
• Distance dependence:
  Force F ∝ \[\frac {1}{r^2}\]
  Electric field E ∝ \[\frac {1}{r^2}\]
  Potential V ∝ \[\frac {1}{r}\]

Calculate the potential at a point P due to a charge of 4 × 10⁻⁷ C located 9 cm away. Hence, obtain the work done in bringing a charge of 2 × 10⁻⁹ C from infinity to point P. State whether the answer depends on the path.

Given:

• Q = 4 × 10⁻⁷ C
• r = 9 cm = 0.09 m
• q = 2 × 10⁻⁹ C (charge brought from infinity)
• \[\frac {1}{4πε_0}\] = 9 × 10⁹ N m²C⁻²

(a) Potential at P:

Compute:

(b) Work done in bringing q from infinity to P:

Path dependence: The work done is independent of the path because the electrostatic field is conservative. Any small displacement can be resolved into a radial component and a perpendicular component; only the radial component contributes to work.

This example uses the fact that every point on the ring is at the same distance (the radius) from the centre, so the ring can be treated like a point charge for potential calculation at the centre. Given a wire bent into a circle of radius R = 10 cm = 10⁻¹ m carrying a uniformly spread charge q = 250 μC = 250 × 10⁻⁶ C, the potential formula for a point charge applies directly since all charge elements are equidistant from the centre.

Here, four different charges sit at the four corners of a square, and potential is calculated at the centre using the superposition principle. Since the square has side \[\sqrt{2}\ \text{m}\], the distance from the centre to each corner (half the diagonal) works out to r = 1 m for all four corners equally.

Because potential is a scalar, the net potential is simply the algebraic sum of individual potentials due to each charge — no vector addition is needed, and since all distances are equal, r can be factored out:

Substituting the charges q₁ = +2 × 10⁻⁹ C, q₂ = +1 × 10⁻⁹ C, q₃ = −2 × 10⁻⁹ C, q₄ = +3 × 10⁻⁹ C:

• Very high electric potentials near charged objects are used in devices like photocopiers, laser printers, and particle accelerators, where charged particles are accelerated by electric fields.
• Lightning can be understood in terms of a huge potential difference between clouds and ground, causing charge to move suddenly through the air.

• Electric potential at a point is the work done per unit positive test charge in bringing it slowly from infinity to that point, against the electric field.
• For a point charge q in air/vacuum:
  V(r) = \[\frac{1}{4\pi\varepsilon_0}\frac{q}{r}\]
• In a medium of dielectric constant K:
  V(r) = \[\frac{1}{4\pi\varepsilon_0K}\frac{q}{r}\]
• Positive charge produces positive potential; negative charge produces negative potential.
• Potential due to a point charge is spherically symmetric and depends only on distance r.
• Distance dependence:
  F ∝ 1/r², E ∝ 1/r², V ∝ 1/r.
• The potential at infinity is taken as zero; only potential differences are physically significant.
• The electrostatic field is conservative, so the work done in moving a charge between two points is path independent.