Final answer:
The charge on the positive electrode of the parallel-plate capacitor is 2.40 nC.
Step-by-step explanation:
Given:
- Area of each electrode, A = 3.0 cm x 3.0 cm = 9 cm² = 0.0009 m²
- Distance between the electrodes, d = 2.2 mm = 0.0022 m
- Electric field strength inside the capacitor, E = 1.0x10⁶ N/C
We know that the electric field between the plates of a parallel-plate capacitor is given by the equation E = V/d, where V is the potential difference between the plates. Rearranging the equation, we have V = E x d. Plugging in the given values, we get:
V = (1.0x10⁶ N/C) x (0.0022 m) = 2200 V
The charge on a capacitor is given by the equation Q = C x V, where Q is the charge, C is the capacitance, and V is the potential difference. Rearranging the equation, we have C = Q/V. Since the capacitance is the same for both electrodes of a parallel-plate capacitor, we can find the charge on one electrode by dividing the total charge by 2. Assuming the charge on both electrodes is the same, we can use the equation C = ε₀A/d to find the capacitance, where ε₀ is the permittivity of free space, A is the area of the electrode, and d is the distance between the electrodes. Rearranging the equation, we have Q = C x V = ε₀A/d x V. Plugging in the given values, we get:
Q = [(8.85x10⁻¹² C²/Nm²) x (0.0009 m²)] / (0.0022 m) x 2200 V = 2.40 nC
Therefore, the charge on the positive electrode is 2.40 nC.