JEE Class main Answered

Numbers written as potential on plates of capacitor is not clearly visible .

Hence procdure of answering the questions are given below

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If one plate is at a potential V₁ and other plate is at a potential V₂ , then electric field intensity E between plates is

E = ( V₁ - V₂ ) / d   ...................................(1)

where d is distance between plates of capacitor.

Force F acting on charge q  kept in the electric field E is

F = q E  ................................(2)

Hence find out the electric field between the capacitor plates using equation (1)

Then calculate the force as given by eq,(2) .

Then rank according to the magnitude of force