A positively charged particle is in the center of a parallel-plate capacitor tha

Question

A positively charged particle is in the center of a parallel-plate capacitor that has charge plusminus Q on its plates. Suppose the distance between the plates is doubled, with the charged particle remaining in the center. Does the force on this particle increase, decrease, or stay the same? Explain. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. If we assume the size of the parallel plates is large compared to the separation, then doubling the plate separation will the field inside the capacitor That means this transformation will the force.

Explanation / Answer

here,

C = area * e0 /d

when the sepration is doubled

the potential difference , V = Q /C

as the charges remains the same

and capacitance is halved

the potential diffrenece is doubled

and electric feild = V/d

so, as potential is doubled and sepration is also doubled

the the electric feild remains the same

so, same electric feild that gives the same force

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