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Work done in moving q coulomb charge along a closed path of length $l$ by an electric field $\vec E$ is given by |
$\int \vec E. \vec{dl}$ $q\vec E.\vec l$ $2q\vec E.\vec l$ Zero |
Zero |
The correct answer is Option (4) → Zero In electrostatics, the electric field E is conservative. This means that work done in moving a charge around a closed loop is zero, because the net change in potential over a closed path is zero. |
