Question

his 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

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.

1. 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)*(1/ ln(b/a)). i. This is a capacitance for a given length L. ii. You should also consider some limiting cases here and discuss. What happens as b approaches a? And what happens as b>>a?

Answer 1