The work done in moving a charge particle of charge 2 × 1 0 − 8 C between two points having potential difference of 36 V is

Given, q = 2 × 1 0 − 8 , Δ V = 36 V Work done, W = q ⋅ Δ V = 2 × 1 0 − 8 × 36 = 72 × 1 0 − 7 V = 0.72 × 1 0 − 6 V = 0.72 μ V

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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2024

PhysicsEasysingle choice

The work done in moving a charge particle of charge $2 \times 1 0^{- 8} C$ between two points having potential difference of 36 V is

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Given, $q = 2 \times 1 0^{- 8}, Δ V = 36 V$ $Work done, W = q \cdot Δ V = 2 \times 1 0^{- 8} \times 36 = 72 \times 1 0^{- 7} V = 0.72 \times 1 0^{- 6} V = 0.72 μ V$

Stream:VITEEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2024

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