Electric Charge Distribution in Cylindrical Capacitors

AI Thread Summary
A potential difference of 160V is applied across two collinear conducting cylinders with radii of 10 cm and 15 cm, and a height of 38 cm. The charge on the cylinders is calculated using the formula Q = Vε0A/d, but an initial attempt yielded an incorrect result of 1.69 x 10^-10 C instead of the expected 8.34 x 10^-9 C. The discussion highlights the importance of using the correct equations for cylindrical capacitors, as the formula initially applied is meant for parallel plate capacitors. The electric field between the cylinders is also a key focus, with participants discussing the need for accurate calculations. Understanding the differences in capacitor types is crucial for solving these problems correctly.
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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.
 
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