Question

In the figure below, two parallel-plate capacitors (with air between the plates) are connected to a battery. Capacitor 1 has a plate area of 1.5 cm² and an electric field (between its plates) of magnitude 2000 V/m. Capacitor 2 has a plate area of 0.70 cm² and an electric field (between its plates) of magnitude 1300 V/m. What is the total charge on the two capacitors?

The answer is not q = ε_o E₁A₁ +  ε_o E₂A₂ = 8.85 e-12 ( 2000 * .015^2 + 1300 * .007^2) = 4.546 e-12 = 4.56 pC

Answer 1

Given data: The area of cross-section of first capacitor is.A1.5cm 10 cm The electric field between the plates of the first capacitor C isE 2000V/m The area of cross-section of second capacitor is.A, 0.7 cm 10 cm The electric field between the plates of the first capacitor C isE-1300V/m The charge stored on capacitor Cis Her capacitance of the capacitor is C , and potential across the cap actor is Ë Since the potential can be expressed asV-Ed, where elecricfieldis E and distance is The capacitance of a parallel plate capacitor is Substitute C and values inQ, Thus, the charge stored on the capacitor Cis, Similarly the charge stored on the second capacitor is The net charge on both capacitors is =(8.85x10-12 Fm)| (2000vmi(1.5x10"m*)-(1300 v/m)(0.70×10"m:) -3.46x10-*c 3.5 pC