A 28.0 nF air-filled parallel plate capacitor is connected in series with a 1.55
ID: 2144968 • Letter: A
Question
A 28.0 nF air-filled parallel plate capacitor is connected in series with a 1.55 k? resistor and a battery of emf 36.0 V.
A 28.0 nF air-filled parallel plate capacitor is connected in series with a 1.55 k? resistor and a battery of emf 36.0 V. Determine the final charge on the capacitor. Once the capacitor is fully charged and there is no current in the circuit, a sheet of plastic with a dielectric constant of 2.30 is introduced in the gap between the plates of the capacitor, completely filling the gap. If the area of each plate is 0.275 m2 what is the electric field inside the plastic immediately after it is inserted (before there is any change in the charge on the plates of the capacitor)? How is the electric field related to the charge on the plate and the area of the plates? How is the electric field affected by the introduction of the dielectric? N/C Take a moment to think about why current starts to flow after the plastic is inserted and apply the loop rule to find a round trip potential difference equation for the circuit just after the plastic is inserted. Determine the initial current through the resistor immediately after the plastic is inserted. Determine the final charge on the positive plate of the capacitor with the dielectric.Explanation / Answer
(a) Final voltage across capacitor is 36V. So, final charge, Q= C1*V1 = 1.008 x 10^-6 C.
(b) Field = Q/Aepsilon = 1.008*10^-6 /0.275*2.3*8.8*10^-12 = 1.81*10^5 N/C
By introduction of dielectric, E.F is decrease by 2.3 times.
(c) Capacitance changes (increases) as you insert the plastic.
Charge on the parallel plates remains the same and so the voltage drops.
Thus, a potential difference is created which results in the flow of current.
If new capacitance is C2, voltage immediately after plastic is introduced is V2, then C1*V1 = C2*V2 (charge conservation).
since C2 = 2.3*C1, V2 = V1/2.3 = 15.7 V.
Instantaneous current = (36V - 15.7V)/1.55k = 13 mA.
(d) Charge on positive plate = charge on negative plate = 2.3*28nF*36V = 2.32 x 10^-6 C.
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