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www.GOALias.blogspot.comPhysicsFIGURE 2.2 Work done on a test charge qby the electrostatic field due to any givencharge configuration is independentof the path, and depends only onits initial and final positions.In other words, the electrostatic potential (V )at any point in a region with electrostatic field isthe work done in bringing a unit positivecharge (without acceleration) from infinity tothat point.The qualifying remarks made earlier regardingpotential energy also apply to the definition ofpotential. To obtain the work done per unit testcharge, we should take an infinitesimal test chargeδq, obtain the work done δW in bringing it frominfinity to the point and determine the ratioδW/δq. Also, the external force at every point ofthe path is to be equal and opposite to theelectrostatic force on the test charge at that point.FIGURE 2.3 Work done in bringing a unitpositive test charge from infinity to thepoint P, against the repulsive force ofcharge Q (Q > 0), is the potential at P due tothe charge Q.2.3 POTENTIAL DUE TO A POINT CHARGEConsider a point charge Q at the origin (Fig. 2.3). For definiteness, take Qto be positive. We wish to determine the potential at any point P withposition vector r from the origin. For that we mustcalculate the work done in bringing a unit positivetest charge from infinity to the point P. For Q > 0,the work done against the repulsive force on thetest charge is positive. Since work done isindependent of the path, we choose a convenientpath – along the radial direction from infinity tothe point P.At some intermediate point P′ on the path, theelectrostatic force on a unit positive charge is∆ QW = − r24 πεr '∆ ′0Q × 1r2 ˆ ′(2.5)4 πεr '0where rˆ′is the unit vector along OP′. Work doneagainst this force from r′ to r′ + ∆r′ is(2.6)The negative sign appears because for ∆r ′ < 0, ∆W is positive . Totalwork done (W) by the external force is obtained by integrating Eq. (2.6)from r′ = ∞ to r′ = r,rrQ Q QW =− ∫ dr′= =2(2.7)4 π ε0r ' 4πε0r′∞ 4πε0r∞This, by definition is the potential at P due to the charge Q54QVr () = 4 π ε r(2.8)0