Exploration 26.5: Capacitance of Concentric Cylinders
Please wait for the animation to
completely load.
Wait for the calculation to finish. This animation shows a coaxial capacitor
with cylindrical geometry:
a very long cylinder (extending into and out of the page) in the center
surrounded by a very long cylindrical shell (position is given in
centimeters, electric field strength is given in newtons/coulomb, and electric potential is
given in volts). The outside shell is grounded, while the inside
shell is at 10 V. You can click-drag to measure the voltage at any
position. Restart.
- Use Gauss's law to show that the magnitude of the radial electric field
between the two conductors for a
cylindrical coaxial capacitor of length L
is E = Q/2πrLε0 = 2kQ/(rL), where Q is the total charge
on the inside (or outside) conductor and r is the distance from the center.
- If L = 1 m, measure the electric field in the region between the two
conductors and determine the charge on the inside (and outside) conductor.
- Use V = -∫ E • dr to show that the potential at any point between the
two conductors is V = (Q/2πLε0) ln(b/r) = (2kQ/L) ln(b/r),
where b is the radius of the outer conductor.
- Given that the potential difference between the two cylinders is 10 V,
verify your answer to (b) and find the charge on each conductor.
- Given, then, that the potential difference between the two conductors is
V = (Q/2πLε0) ln(b/a) = (2Qk/L) ln(b/a)–b is the radius of the outer shell
and a is the radius of the inner cylinder–show that the capacitance of this
capacitor is (2πLε0)/ln(b/a) = (L/2k)/ln(b/a).
- What is the capacitance (numerical value) of this capacitor?
Exploration authored by Anne J. Cox.
Script authored by Mario Belloni, Wolfgang Christian and Anne J. Cox.
© 2004 by Prentice-Hall, Inc. A Pearson Company