Calculation of Capacitance and Applied Potential Difference
A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 15.5 pC. The inner cylinder has a radius of 0.600 mm, the outer one has a radius of 5.20 mm, and the length of each cylinder is 25.0 cm.
(a) What is the capacitance?
We need to calculate the capacitance using the formula of a capacitor:
\[C = \frac{2\pi\epsilon_{0}l}{\ln\frac{b}{a}}\]
Given:
Charge = 15.5 pC
Inner Radius = 0.600 mm
Outer Radius = 5.20 mm
Length = 25.0 cm
Putting the values into the formula:
\[C = \frac{2\pi \times 8.85 \times 10^{-12} \times 25.0 \times 10^{-2}}{\ln\frac{5.20}{0.600}}\]
\[C = 6.43 \times 10^{-12} F\]
Therefore, the capacitance is \[6.43 \times 10^{-12} F\].
(b) What applied potential difference is necessary to produce these charges on the cylinders?
To find the potential difference required, we use the formula of charge:
\[q = CV\]
\[V = \frac{q}{C}\]
Plugging in the values:
\[V = \frac{15.5 \times 10^{-12}}{6.43 \times 10^{-12}}\]
\[V = 2.41 V\]
Thus, the potential difference necessary to produce these charges on the cylinders is 2.41 V.
What are the components of the capacitor in this scenario?
The capacitor consists of two hollow, coaxial, iron cylinders with different charges applied to them. The inner cylinder is negatively charged, while the outer cylinder is positively charged.