2.3 Potential due to a Point Charge

Consider a point charge Q at the origin (Fig.2.3).

FIGURE 2.3: Work done in bringing a unit positive test charge from infinity to the point P, against the repulsive force of charge Q (Q > 0), is the potential at P due to the charge Q.

For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r vector from the origin. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path - along the radial direction from infinity to the point P.

At some intermediate point P′ on the path, the electrostatic force on a unit positive charge is


[Q/ (4πε0 r′^2)] r′ cap .. .. (2.5)


where r′ cap is the unit vector along oP′.


Work done against this force from r′ vector to r′ vector + Δr′ vector is


ΔW = − [Q/ (4πε0 r′^2)] Δr′ .. .. (2.6)

The negative sign appears because for Δr′ > 0, ΔW is negative. Total work done (W) by the external force is obtained by integrating Equation (2.6) from r′ = ∞ to r′ = r,

W = − ∫ from ∞ to r (Q/4πε0 r′^2) × dr′


= Q/4πε0 r′| from ∞ to r


= Q/4πε0 r .. .. (2.7)

This, by definition is the potential at P due to the point charge Q at the origin.

V(r) = Q/4πε0 r .. .. (2.8)

Equation (2.8) is true for any sign of the charge Q; though we considered Q > 0 in its derivation.

For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge from infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are in the same direction.]

Finally, we note that Equation (2.8) is consistent with the choice that potential at infinity be zero.

Figure (2.4) shows how the electrostatic potential (∝ 1/r) and the electrostatic field (∝ 1/r^2) varies with r.

FIGURE 2.4 Variation of potential V with r [in units of (Q/4 πε0) m^−1] (blue curve) and field with r [in units of (Q/4πε0) m^−2] (black curve) for a point charge Q.

Example 2.1

(a) Calculate the potential at a point P due to a charge of 4 × 10^−7C located 9 cm away.


(b) Hence obtain the work done in bringing a charge of 2 × 10^−9 C from infinity to the point P. Does the answer depend on the path along which the charge is brought?

Solution:


(a) V = 1/ (4πε0) × (Q/r)


= 9 × 10^9 Nm^2 C^−2 × (4 × 10^−7C/0.09 m)


= 4 × 10^4 V

(b) W = qV = 2 × 10^−9C × 4 × 10^4V


= 8 × 10^−5 J

No, work done will be path independent. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r vector and another perpendicular to r vector. The work done corresponding to the later will be zero.