Show that the work done in an electric field is independent of path
Answers
Answered by
119
Work done in an electric field by transporting an object having an electric charge q coulombs from a point having a potential V1 volts to a point having potential V2 is given by
W = q (V2 - V1)
The work done is also defined as the force * displacement.
![W= \int\limits^{P_2}_{P_1} {F} \, ds\\\\=\int\limits^{P_2}_{P_1} {[q*E(x,y)]} \, ds\\\\=q*\int\limits^{P_2}_{P_1} {E(x,y)} \, ds\\\\=q*[V]_{V_1}^{V_2}=q*[V_2-V_1]\\\\as,\ E(x,y)=\frac{dV}{ds}\\Electric\ field\ at\ a\ point\ (x,y)=space\ gradient\ of\ electric\ potential](https://tex.z-dn.net/?f=W%3D+%5Cint%5Climits%5E%7BP_2%7D_%7BP_1%7D+%7BF%7D+%5C%2C+ds%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7BP_2%7D_%7BP_1%7D+%7B%5Bq%2AE%28x%2Cy%29%5D%7D+%5C%2C+ds%5C%5C%5C%5C%3Dq%2A%5Cint%5Climits%5E%7BP_2%7D_%7BP_1%7D+%7BE%28x%2Cy%29%7D+%5C%2C+ds%5C%5C%5C%5C%3Dq%2A%5BV%5D_%7BV_1%7D%5E%7BV_2%7D%3Dq%2A%5BV_2-V_1%5D%5C%5C%5C%5Cas%2C%5C+E%28x%2Cy%29%3D%5Cfrac%7BdV%7D%7Bds%7D%5C%5CElectric%5C+field%5C+at%5C+a%5C+point%5C+%28x%2Cy%29%3Dspace%5C+gradient%5C+of%5C+electric%5C+potential "W= \int\limits^{P_2}_{P_1} {F} \, ds\\\\=\int\limits^{P_2}_{P_1} {[q*E(x,y)]} \, ds\\\\=q*\int\limits^{P_2}_{P_1} {E(x,y)} \, ds\\\\=q*[V]_{V_1}^{V_2}=q*[V_2-V_1]\\\\as,\ E(x,y)=\frac{dV}{ds}\\Electric\ field\ at\ a\ point\ (x,y)=space\ gradient\ of\ electric\ potential")
So no matter whichever path S is taken from point P1 to point P2, the integral will have only one value.
W = q (V2 - V1)
The work done is also defined as the force * displacement.
So no matter whichever path S is taken from point P1 to point P2, the integral will have only one value.
Answered by
1
Explanation:
This Proves that Work Done by an electric charge is a State function i.e. it depends on the final and initial state of the charge and not on the path is taken.
Similar questions