Once the plates are fully charged, the battery is removied, a charge of q is mai

Question

Once the plates are fully charged, the battery is removied, a charge of q is maintained. Given that the plate dimensions are .5mX.5m, what is the force on the conductor after length =0.25m has been inserted into the capacitor? Assume the charge on the plates remain constant.

Once the plates are fully charged, the battery is removed, a charge of q is maintained. Given that the plate dimensions are .5mX.5m, what is the force on the conductor after length =0.25m has been inserted into the capacitor? Assume the charge on the plates remain constant.

Explanation / Answer

force between capacitor plates is F =q^2/2?0A

initial area of the plates is A =0.50m*0.50m

=0.25 m^2

after a length x =0.25 m has been inserted the area of the each plate is A1 =0.50*0.25

=0.125 m^2

force between two plates after a length x =0.25 m has been inserted is F1=q^2/2*0.125?0

=4q^2/?0

if area is consider in units of cm^2

we get the answer for force between two plates after a length x =0.25 m has been inserted is

F1=0.04q^2/?0 is correct answer

hence the answer is A

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