Each plate of a parallel-plate capacitor is a squarewith side length,r, and the

Question

Each plate of a parallel-plate capacitor is a squarewith side length,r, and the plates are separated by adistance, d. The capacitor is connected to a source of voltage V. A plastic slab of thickness, d , and dielectric K isinserted slowly between the plates over the time period ,t, untilthe slab is squarely between the plates. While the slab is beinginserted, a current urns through the battery/capacitorcircuit.
Assuming the dielectric is inserted at a constant rate, findthe current (I) as the slab in inserted. Please use any/all the variables: V,K r, d,t,0 Each plate of a parallel-plate capacitor is a squarewith side length,r, and the plates are separated by adistance, d. The capacitor is connected to a source of voltage V. A plastic slab of thickness, d , and dielectric K isinserted slowly between the plates over the time period ,t, untilthe slab is squarely between the plates. While the slab is beinginserted, a current urns through the battery/capacitorcircuit.
Assuming the dielectric is inserted at a constant rate, findthe current (I) as the slab in inserted. Please use any/all the variables: V,K r, d,t,0

Explanation / Answer

Here current flows because potential remains constant , so whencapacitance changes charge stored in capacitor change ,and hencecurrent flows Here initial capacitance = A/d Here A = area = r2 Putting C = r2/d Now , initial charge = CV = Vr2/d Finally , after insertion of plate ,final capacitance=r2 /( (d-t) + kt) = kr2/d So final charge = Vkr2 /d As slab is put with uniform velocity , so average current = Totalcharge /time                      = (Vkr2 /d -Vr2 /d ) /t =Vor2(k-1)/td

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