A cylindrical capacitor has two co-axial cyli | vedclass

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(D) The capacitance $C$ of a cylindrical capacitor is given by the formula: $C = \frac{2 \pi \varepsilon_0 l}{\ln(b/a)}$ where $l = 20 \, cm = 0.2 \,

Vedclass · Accepted Answer

$\frac{0.01 \ln 2}{4 \pi \varepsilon_0} \, V$

Vedclass · Answer

(D) The capacitance $C$ of a cylindrical capacitor is given by the formula: $C = \frac{2 \pi \varepsilon_0 l}{\ln(b/a)}$ where $l = 20 \, cm = 0.2 \, m$,$a = r$,and $b = 2r$. Substituting the values: $C = \frac{2 \pi \varepsilon_0 (0.2)}{\ln(2r/r)} = \frac{0.4 \pi \varepsilon_0}{\ln 2}$. The potential difference $\Delta V$ is given by $\Delta V = \frac{Q}{C}$. Given $Q = 10 \, \mu C = 10 	imes 10^{-6} \, C$. $\Delta V = \frac{10 	imes 10^{-6}}{\frac{0.4 \pi \varepsilon_0}{\ln 2}} = \frac{10 	imes 10^{-6} \ln 2}{0.4 \pi \varepsilon_0}$. $\Delta V = \frac{10^{-5} \ln 2}{0.4 \pi \varepsilon_0} = \frac{10^{-4} \ln 2}{4 \pi \varepsilon_0} = \frac{0.0001 \ln 2}{4 \pi \varepsilon_0} \, V$. Note: The provided options seem to have unit discrepancies or scaling errors. Based on the calculation,the magnitude is $\frac{10^{-4} \ln 2}{4 \pi \varepsilon_0} \, V$.

$A$ cylindrical capacitor has two co-axial cylinders of length $20 \, cm$ and radii $2r$ and $r$. The inner cylinder is given a charge of $10 \, \mu C$ and the outer cylinder a charge of $-10 \, \mu C$. The potential difference between the two cylinders will be:

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