Question

Show that the work done in an electric field is independent of path.

Answer 1

Work done in an electric field by transporting an object having an electric charge $q$ coulombs from a point having a potential $V_1$ volts to a point having potential $V_2$ is given by-

$W=q(V_2-V_1)$

The work done is also defined as the $force\times displacement$.

$W=\underset{P_1}{\overset{P_2}{\int }}Fds=\underset{P_1}{\overset{P_2}{\int }}[q\times E(x,y\left)\right]ds=q\times \underset{P_1}{\overset{P_2}{\int }}E(x,y)ds=q\times [V{]}_{V_1}^{V_2}=q\times [V_2-V_1]as,E(x,y)=\frac{dV}{ds}Electricfieldatapoint(x,y)=spacegradientofelectricpotential.$

So no matter whichever path S is taken from point $P_{1}topoint{P}_2$, the integral will have only one value.