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Derive an expression for the electric potentialin a electric field of positive point charge at distance r. |
Answer» Solution :Consider a point charge Q at the ORIGIN as SHOWN in figure. The work done in bringing a unit positive test charge from infinity to the point P. For Q which is positive the work done against the repulsive force on the test charge is positive. Since the 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, `F=(kQxx1)/((r)^(2))hatr` Where f is the unit vector along OP. Work done against this force from r. to r +` Deltar` is `DeltaW= -(kQ)/((r)^(2)). Deltar`[ From W = F r cos`theta]` Here `Deltar lt 0 ` so `DeltaW gt 0` in this formula . Total work done by the external force is obtained by INTEGRATING above equation from `r = prop ` to r = r . `:. -int_(oo)^(r)(kQ)/((r)^(2)).dr [underset(Deltar rarr0)lim=dr]` `:. W =- kQ [-(1)/(r)]_(oo)^(r) = (kQ)/(r)` `:. W= (Q)/(4pi in_(0)r)` From definition of ELECTRIC potential `V(r) = (W)/(q)=(W)/(I) [ because q=1C]` `:. V(r) = W= (kQ)/(r)` |
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