Electric Charge Distribution in Cylindrical Capacitors

[kgal]

Homework Statement

A potential difference of 160V is applied across two col-linear conducting cylinders. the radius of the outer cylinder is 15 cm, the radius of the inner cylinder is 10 cm, the height of the two cylinders is 38 cm.
a. How much charge is applied to each of the cylinders?
b. What is the magnitude of the electric field between the two cylinders? (magnitudes inside inner and outer cylinder)

Homework Equations

V = Ed = Qd/ε0A

The Attempt at a Solution

a.
V = Qd/ε0A
Q = Vε0A/d = [Vε0(2pi(r1 - r2)^2*h] / d ///(d and one of the (r1 - r2)^2 cancel)
Q = (160)(8.85*10^-12)(2pi)(0.05)(0.38) = 1.69*10^-10 C.
The answer is supposed to be 8.34 *10^-9 C.

 

[ehild]

The Attempt at a Solution

a.
V = Qd/ε0A

The formula applies to parallel plate capacitors. This is cylindrical one.

See:http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html#c2


ehild