Two 14-cm-diameter electrodes 0.46 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 18 V battery.
The correct answer and explanation is:
To calculate the capacitance of a parallel-plate capacitor, the formula is given by: C=ε0⋅AdC = \frac{\varepsilon_0 \cdot A}{d}
where:
- CC is the capacitance in farads (F),
- ε0\varepsilon_0 is the vacuum permittivity (8.85×10−12 F/m8.85 \times 10^{-12} \, \text{F/m}),
- AA is the area of the plate in square meters (m²),
- dd is the separation between the plates in meters (m).
Step 1: Calculate the area of the electrode
The electrodes are circular, so the area AA of one electrode is calculated using the formula for the area of a circle: A=πr2A = \pi r^2
The diameter of each electrode is given as 14 cm, so the radius rr is half of that: r=14 cm2=7 cm=0.07 mr = \frac{14 \, \text{cm}}{2} = 7 \, \text{cm} = 0.07 \, \text{m}
Now, calculate the area: A=π×(0.07)2=π×0.0049 m2≈0.0154 m2A = \pi \times (0.07)^2 = \pi \times 0.0049 \, \text{m}^2 \approx 0.0154 \, \text{m}^2
Step 2: Convert the separation distance between the plates
The separation between the plates is 0.46 cm, which needs to be converted to meters: d=0.46 cm=0.0046 md = 0.46 \, \text{cm} = 0.0046 \, \text{m}
Step 3: Calculate the capacitance
Substitute the values of ε0\varepsilon_0, AA, and dd into the capacitance formula: C=(8.85×10−12)×0.01540.0046C = \frac{(8.85 \times 10^{-12}) \times 0.0154}{0.0046} C≈1.36×10−130.0046≈2.96×10−11 FC \approx \frac{1.36 \times 10^{-13}}{0.0046} \approx 2.96 \times 10^{-11} \, \text{F}
Thus, the capacitance of the parallel-plate capacitor is approximately 2.96×10−112.96 \times 10^{-11} farads or 29.6 picofarads (pF).
Step 4: Determine the charge on the capacitor
The charge stored on the capacitor can be calculated using the formula: Q=C⋅VQ = C \cdot V
where VV is the voltage across the capacitor, which is 18 V. Substituting the values: Q=(2.96×10−11 F)×(18 V)=5.33×10−10 CQ = (2.96 \times 10^{-11} \, \text{F}) \times (18 \, \text{V}) = 5.33 \times 10^{-10} \, \text{C}
Thus, the charge stored on the capacitor is approximately 5.33×10−105.33 \times 10^{-10} coulombs or 533 nanocoulombs (nC).