Question Detail
Question
A parallel plate capacitor has an area of 2000 cm2, and the plates are 1 cm apart. The original potential difference between them, V0, is 3000 volts, and it decreases to 1000 volts when a sheet of dielectric is inserted between the plates. Compute the induced charge Qind on each face of the dielectric. (Take ε0 = 8.854 × 10-12 C2/Nm2)
Correct Answer
Option D is the correct answer.
Detailed Explanation
First, calculate the capacitance without the dielectric (C0):Area (A) = 2000 cm2 = 2000 × 10-4 m2 = 0.2 m2Distance (d) = 1 cm = 0.01 mC0 = ε0A/d = (8.854 × 10-12 C2/Nm2) * (0.2 m2) / (0.01 m) = 1.7708 × 10-10 FThe original charge (Qfree) on the capacitor (before dielectric insertion, or if isolated after charging) is:Qfree = C0V0 = (1.7708 × 10-10 F) * (3000 V) = 5.3124 × 10-7 CWhen the dielectric is inserted, the voltage drops from 3000 V to 1000 V. This implies the capacitor was disconnected from the source, so the free charge Qfree remains constant.The dielectric constant (k) can be found from the voltage ratio:k = V0 / Vdielectric = 3000 V / 1000 V = 3The induced charge (Qind) on the dielectric faces is related to the free charge by:Qind = Qfree * (1 - 1/k)Qind = (5.3124 × 10-7 C) * (1 - 1/3)Qind = (5.3124 × 10-7 C) * (2/3)Qind ≈ 3.5416 × 10-7 CConverting this to the options' format:Qind = 35.416 × 10-8 CThe closest option is 35.4 × 10-8 C.
Hint
First, calculate the original charge on the capacitor and the dielectric constant. Then, use the relationship between free charge, induced charge, and the dielectric constant.