Answer
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Hint: A force is said to be conserved if the work done by the force in moving a particle form one point to another point depends totally only on the initial and final points. The force acting on the particle doesn’t depend on the path followed.
Complete step by step answer:
Let us first assume two points a and b in an electric field due to charge q. Some amount of work is done to move a test charge from one point to another point along a closed path. We need to find the line integral of the electric field along a to b and then b to a and find the relation between them. If they add you zero, the work done is independent of the path and depends only on the end points of a and b.
Line integral of electric field along a to b will be,
$\int\limits_{A}^{B}{E.dl}$
Similarly, the line integral of electric field along b to a path will be,
$\int\limits_{B}^{A}{E.dl}$
The line integral along the closed path will be,
$\int\limits_{B}^{A}{E.dl}+\int\limits_{A}^{B}{E.dl}$
But we know,
$\int\limits_{B}^{A}{E.dl}=-\int\limits_{A}^{B}{E.dl}$
Hence, the line integral of a closed path will be equal to zero.
Therefore, we can conclude that the thaw work done by the electric field is indecent of the path and depends only on the end points a and b of integration. We proved this by solving the line integral and gaining the line integral as zero along a long path.
Note:
The electric field is conservative in nature. We can prove this by proving that the work done by an electric field depends only on the starting and the ending points of the cycle but not on the path taken by the electric field.
Complete step by step answer:
Let us first assume two points a and b in an electric field due to charge q. Some amount of work is done to move a test charge from one point to another point along a closed path. We need to find the line integral of the electric field along a to b and then b to a and find the relation between them. If they add you zero, the work done is independent of the path and depends only on the end points of a and b.
Line integral of electric field along a to b will be,
$\int\limits_{A}^{B}{E.dl}$
Similarly, the line integral of electric field along b to a path will be,
$\int\limits_{B}^{A}{E.dl}$
The line integral along the closed path will be,
$\int\limits_{B}^{A}{E.dl}+\int\limits_{A}^{B}{E.dl}$
But we know,
$\int\limits_{B}^{A}{E.dl}=-\int\limits_{A}^{B}{E.dl}$
Hence, the line integral of a closed path will be equal to zero.
Therefore, we can conclude that the thaw work done by the electric field is indecent of the path and depends only on the end points a and b of integration. We proved this by solving the line integral and gaining the line integral as zero along a long path.
Note:
The electric field is conservative in nature. We can prove this by proving that the work done by an electric field depends only on the starting and the ending points of the cycle but not on the path taken by the electric field.
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