A parallel-plate air capacitor of area A = 10.5 cm2 and plate separation of d =

Question

A parallel-plate air capacitor of area A = 10.5 cm2 and plate separation of d = 3.80 mm is charged by a battery to a voltage of 53.0 V. What is the charge on the capacitor? Incompatible units. No conversion found between "*f" and the required units. If a dielectric material with k = 4.60 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate? Calculate the change in capacitance, and then the change in charge.

Explanation / Answer

the equaiton for parallel plate capacitor is

C = e_0 A/d

charge on the capacitor is

Q = CV=  e_0 A/d V = 8.85 * 10 ^-12 ( 10.5 * 10^-4 m^2/3.80 * 10 ^-3 m) 53 V=129.6 * 10 ^-12 C

if dielectric is inserted between plates the additional charge is

Q' = CV= ke_0 A/d V = (4.6)8.85 * 10 ^-12 ( 10.5 * 10^-4 m^2/3.80 * 10 ^-3 m) 53 V=518.42 * 10 ^-12 C

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