Answer :
Answer:
(A) Capacitance per unit length = [tex]4.02 \times 10^{-10}[/tex]
(B) The magnitude of charge on both conductor is [tex]Q = 4.22 \times 10^{-19}[/tex] C and the sign of charge on inner conductor is [tex]+Q[/tex] and the sign on outer conductor is [tex]-Q[/tex]
Explanation:
Given :
Radius of inner part of conductor [tex](R_{1})[/tex] = [tex]2.7 \times 10^{-3}[/tex] m
Radius of outer part of conductor [tex](R_{2})[/tex] = [tex]3.1 \times 10^{-3}[/tex] m
The length of the capacitor [tex](l)[/tex] = [tex]3 \times 10^{-3}[/tex] m
(A)
Capacitance is purely geometrical property. It depends only on length, radius of conductor.
From the formula of cylindrical capacitor,
[tex]C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }[/tex]
Where, [tex]\epsilon_{o} = 8.85 \times 10^{-12}[/tex]
But we need capacitance per unit length so,
[tex]\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }[/tex]
capacitance per unit length = [tex]\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}[/tex]
(B)
The charge on both conductors is given by,
[tex]Q = C \Delta V[/tex]
Where, C = capacitance of cylindrical capacitor and value of [tex]C = 12.06 \times 10^{-13}[/tex] F, [tex]\Delta V = 350 \times 10^{-3}[/tex] V
∴ [tex]Q = 4.22 \times 10^{-19}[/tex] C
The magnitude of charge on both conductor is same as above but the sign of charge is different.
Charge on inner conductor is [tex]+Q[/tex] and Charge on outer conductor is [tex]-Q[/tex].