Two long, straight wires are located parallel to each other in space at a separation of ‘b’ between their axes. Each wire has cross-sectional radius equal to ‘a’. The mutual capacitance of this system per unit length of the wires under the condition a<<b is nπε0ln(ba) Then ‘n’ is

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Two long, straight wires are located parallel to each other in space at a separation of ‘ $b$ ’ between their axes. Each wire has cross-sectional radius equal to ‘ $a$ ’. The mutual capacitance of this system per unit length of the wires under the condition $a < < b$ is $\frac{n π ε_{0}}{\ln (\frac{b}{a})}$ Then ‘ $n$ ’ is

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Detailed Solution

$E = \frac{λ}{2 π ε_{0} x} + \frac{λ}{2 π ε_{0}} \frac{1}{b - x}$

$- d V = E d x = - [\frac{λ}{2 π ε_{0} x} + \frac{λ}{2 π ε_{0}} \frac{1}{b - x}] d x$

$V = - \frac{λ}{2 π ε_{0}} \int_{a}^{b - a} [\frac{1}{x} + \frac{1}{b - x}] d x$

$V = - \frac{λ}{2 π ε_{0}} {[\ln x - \ln (b - x)]}_{a}^{b - a}$

$V = - \frac{λ}{2 π ε_{0}} [2 \ln (b - a) - 2 \ln a]$

$V = - \frac{λ}{2 π ε_{0}} [2 \ln (b) - 2 \ln a]$ (since $b > > a$ )

$V = - \frac{λ}{π ε_{0}} \ln (\frac{b}{a})$

Capacitance per unit length,

$C = \frac{q}{|V|} \frac{1}{d x} = \frac{λ d x}{|V|} \frac{1}{d x} = \frac{λ}{|V|} = \frac{λ}{\frac{λ}{π ε_{0}} \ln (\frac{b}{a})} = \frac{π ε_{0}}{\ln (\frac{b}{a})}$