A parallel-plate capacitor, with air between the plates, is connected to a batte

Question

A parallel-plate capacitor, with air between the plates, is connected to a battery. The battery establishes a potential difference between the plates by placing charge of magnitude q=2.15 × 10-6 C on each plate. The space between the plates is then filled with a dielectric material, with dielectric constant k = 4.82. What must the magnitude of the charge in micro coulombs on each capacitor plate now be, to produce the same potential difference between the plates as before? Express the answer with two decimal places.

Explanation / Answer

Since capacitance C is C = K A/d where K= ek e- permitivity of free space k- relative permitivity A -area of the plates d- distance between the plates then C1/C2 = [ e A/d ] /[ ek A/d ] and C1/C2 = 1/k Also charge Q is related to capacitance C and voltage V accross its terminal as Q=CV Since V=Q/C V1=V2 we have Q1/C1 = Q2/C2 and then Q1/Q2 = C1/C2 = 1/k then Q2 = k Q1 Q2=4.82 x Q2 = k Q1 Q2=4.82 x 2.15 x 10^-6 Q2=10.36 x 10^-6 C

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