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lam58
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The question asks to find the potential of the ungrounded outer conducting sheath of a coaxial cable when the inner conductor has a voltage of [tex]\frac{220}{\sqrt{3}} kV[/tex].
Coaxial conditions:
i) Central copper conductor of diameter 4.8 cm.
ii) Insulating layer of XPLE of thickness 2.3 cm and relative permittivity k = 2.2.
iii) Conducting lead sheath of thickness 2.9 mm.
iv) Insulating HDPE layer of thickness 5 mm and k = 2.4.
v) Outer coating negligible thickness in contact with soil.
My attempt:
I found the XPLE to have a capacitance of 310.70x10^-12 F/m.
To find the potential I assumed the charge between both conductors would equal 0, thus the charge on the inner conductor = -'ve charge on sheath. Hence I can use Q = CV to find the charge:
[tex] 310.73x10^{-12} * \frac{220}{\sqrt{3}} = 3.95x10^{-5} C [/tex]
Then to find potential between the two I use:
[tex] v = \frac{-Q}{2\pi\epsilon_0\epsilon_r}.ln(b/a)[/tex]
[tex]\Rightarrow v = \frac{-Q}{2\pi * 8.8x10^{-12} * 2.4}.ln(0.0789/0.0739) = -19.5 kV[/tex]
Which implies the potential on the lead sheath = [tex]\frac{220}{\sqrt{3}} kV + -19.5kV = 107.5kV[/tex]
Have I done this right?
Coaxial conditions:
i) Central copper conductor of diameter 4.8 cm.
ii) Insulating layer of XPLE of thickness 2.3 cm and relative permittivity k = 2.2.
iii) Conducting lead sheath of thickness 2.9 mm.
iv) Insulating HDPE layer of thickness 5 mm and k = 2.4.
v) Outer coating negligible thickness in contact with soil.
My attempt:
I found the XPLE to have a capacitance of 310.70x10^-12 F/m.
To find the potential I assumed the charge between both conductors would equal 0, thus the charge on the inner conductor = -'ve charge on sheath. Hence I can use Q = CV to find the charge:
[tex] 310.73x10^{-12} * \frac{220}{\sqrt{3}} = 3.95x10^{-5} C [/tex]
Then to find potential between the two I use:
[tex] v = \frac{-Q}{2\pi\epsilon_0\epsilon_r}.ln(b/a)[/tex]
[tex]\Rightarrow v = \frac{-Q}{2\pi * 8.8x10^{-12} * 2.4}.ln(0.0789/0.0739) = -19.5 kV[/tex]
Which implies the potential on the lead sheath = [tex]\frac{220}{\sqrt{3}} kV + -19.5kV = 107.5kV[/tex]
Have I done this right?