Four capacitors of 10 pF are connected in parallel with two of the cap...
Explanation:
Calculating the charge stored on capacitors:
- The charge stored on a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage applied.
- Since all four capacitors are connected in parallel, the total capacitance of the circuit is the sum of the individual capacitances, which is 40 pF.
Calculating the charge stored on Mylar capacitors:
- For the Mylar capacitors, the dielectric constant (κ) is 3, so the capacitance is increased by a factor of 3.
- Therefore, the capacitance of each Mylar capacitor is 3 * 10 pF = 30 pF.
- The total capacitance of the two Mylar capacitors is 60 pF.
- Using the formula Q = CV, the charge stored on the Mylar capacitors is Q = 60 pF * 40 V = 2400 pC.
Calculating the charge stored on air capacitors:
- For the air capacitors, the capacitance remains at 10 pF each.
- The total capacitance of the two air capacitors is 20 pF.
- Using the formula Q = CV, the charge stored on the air capacitors is Q = 20 pF * 40 V = 800 pC.
Conclusion:
- Comparing the charge stored on the Mylar capacitors (2400 pC) with the charge stored on the air capacitors (800 pC), we can see that the Mylar capacitors store 3 times as much charge as the air capacitors.
- Therefore, the correct answer is option B.
Four capacitors of 10 pF are connected in parallel with two of the cap...
- Dielectrics help increase the capacitance of the capacitor, allowing additional charge to be stored on the capacitor.
- Capacitance increases by the dielectric constant as: Cnew = C*K. Given C = ε0A/d and Q = CV, we can have the modified Q = Kε0 AV/d for a capacitor with a dielectric.
- The Mylar capacitors store 3 times as much charge as the air capacitors.
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