A car starts from rest and accelerates uniformly at 2 m/s² for 10 s. Distance covered in 10th second is MCQ with Answer and Explanation
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The work done in moving a charge q along an equipotential surface is:
Correct Answer: Option A
Explanation:
Equipotential surface has constant potential V everywhere. Work done W = qΔV; since ΔV = 0 along equipotential, W = 0. This holds for any path on the surface. Electric field is perpendicular to equipotential surfaces, so no work is done moving charge along them. Memory aid: 'Equipotential: ΔV=0 ⇒ W=0; field lines ⊥ equipotentials'. This electrostatics concept is frequently tested in competitive exams. Always link work to potential difference, not absolute potential; path independence is key for conservative fields.
Equipotential surface has constant potential V everywhere. Work done W = qΔV; since ΔV = 0 along equipotential, W = 0. This holds for any path on the surface. Electric field is perpendicular to equipotential surfaces, so no work is done moving charge along them. Memory aid: 'Equipotential: ΔV=0 ⇒ W=0; field lines ⊥ equipotentials'. This electrostatics concept is frequently tested in competitive exams. Always link work to potential difference, not absolute potential; path independence is key for conservative fields.
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