HW16Sol - d = o d E / dt = (8.85x10-12 C 2 / Nm 2 )...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Jim Guinn’s PHYS2212 Assignment #16 Solutions 1. A parallel-plate capacitor in air has circular plates of radius 2.30cm separated by 1.10mm . Charge is flowing onto the upper plate and off the lower plate at a rate of 5.00A . a) Find the electric field between the plates as a function of time (assume Q = 0 at t = 0) . b) Determine the displacement current between the plates and show that it is 5.00A . a) The charge on the capacitor is given by Q = I t = 5.00A t , and the electric field between the plates is given by E = Q / ε o A = Q / ε o π r 2 = (5.00A t) / [(8.85x10 -12 C 2 / Nm 2 ) π (0.0230m) 2 ] = E = 3.40x10 14 t V / m s . b) The electric flux is given by Φ E = E A = (3.40x10 14 t V / m s) π (0.0230m) 2 = 5.65x10 11 t V m / s , and the displacement current is given by I
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d = o d E / dt = (8.85x10-12 C 2 / Nm 2 ) (5.65x10 11 V m / s) = 5.00 C/s = 5.00A . (Hooray!) 2. In a region of space the electric field varies according to = (0.0500N/C) sin(2.00x10 3 t / sec) . Find the maximum displacement current through a 1.00m 2 area in the y-z plane . For an area in the y-z plane we define the area vector = 1.00m 2 , so E = = (0.0500N/C) sin(2.00x10 3 t / sec) (1.00 m 2 ) = (0.0500N m 2 /C) sin(2.00x10 3 t / sec). The displacement current is then I d = o d E / dt = (8.85x10-12 C 2 / Nm 2 ) (0.0500N m 2 /C) (2.00x10 3 /sec) cos(2.00x10 3 t / sec) = 8.85x10-10 A cos(2.00x10 3 t / sec) . This has a maximum value when cos(2.00x10 3 t / sec) = 1 , so I d, max = 8.85x10-10 A ....
View Full Document

Ask a homework question - tutors are online