A Capacitor: Calculation of Capacitance and Applied Potential Difference
Calculation of Capacitance and Applied Potential Difference in a Capacitor
What is the capacitance?
To calculate the capacitance, we use the formula: \[C = \frac{2 \pi \epsilon_o L}{\ln(b/a)}\] where \(a\) is the radius of the inner cylinder, \(b\) is the radius of the outer cylinder, and \(L\) is the length of the cylinder. Given values are: \(a = 0.550 mm = 5.5 \times 10^{-4} m\), \(b = 4.00 mm = 4.0 \times 10^{-3} m\), and \(L = 15.0 cm = 0.150 m\). Substituting the values: \[C = \frac{2 \times \pi \times 8.854 \times 10^{-12} \times 0.15}{\ln(4.00/0.550)}\] \[C = 4.2057 \times 10^{-12} F\] Therefore, the capacitance is \(4.2057 \times 10^{-12} F\) or \(4.2057 \text{ pF}\).What applied potential difference is necessary to produce these charges on the cylinders?
The relationship between charge and potential difference in a capacitor is given by: \[Q = CV\] Given that the charge \(Q\) is 11.5 pC and the capacitance \(C\) is \(4.2057 \times 10^{-12} F\), we can calculate the potential difference \(V\): \[V = \frac{11.5}{4.2057}\] \[V = 2.7344V\] Hence, the applied potential difference necessary to produce the charges on the cylinders is 2.7344V.What are the formula and values used to calculate the capacitance in the given scenario?
The formula used to calculate the capacitance of the cylindrical capacitor in this scenario is \(C = \frac{2 \pi \epsilon_o L}{\ln(b/a)}\), where \(a\) is the radius of the inner cylinder, \(b\) is the radius of the outer cylinder, and \(L\) is the length of the cylinder. The values given for this calculation are \(a = 0.550 mm\), \(b = 4.00 mm\), and \(L = 15.0 cm\).