2306_sol test 2 form A f07

2306_sol test 2 form A f07 - Physics 2306 Fall 2007 Second...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 2306 Fall 2007 Second Exam Form A 1-4) A solid insulating sphere (radius = r a ) has a total charge + 3q distributed uniformly throughout its volume. A hollow conducting sphere is concentric with the solid sphere and has an inner radius r b and an outer radius r c where r a < r b < r c . A negative charge 2q is on the hollow conducting sphere. 1) What is the charge density on the inner surface of the hollow conducting sphere? A. q/4 π r b 2 B. 2q/4 π r b 2 C. 3q/4 π r b 2 D. 4q/4 π r b 2 E. q/4 π r b 2 F. 2q/4 π r b 2 G. 3q/4 π r b 2 H. 4q/4 π r b 2 2) What is the charge density on the outer surface of the hollow conducting sphere? A. q/4 π r c 2 The enclosed charge within a sphere of radius just greater than r b must be zero so that the electric field be zero within the hollow conductor (Gauss’s law). Thus, the charge on the inner surface r b must be 3q. The charge density is charge/area. The charge on the inner surface of the hollow conductor is 3q. The total charge on the hollow conductor is 2q. Charge on the outer surface equals total charge minus charge on the inner surface = +q. Charge density equals charge on the surface/surface area.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
B. 2q/4 π r c 2 C. 3q/4 π r c 2 D. 4q/4 π r c 2 E. q/4 π r c 2 F. 2q/4 π r c 2 G. 3q/4 π r c 2 H. 4q/4 π r c 2 3) What is the electric potential difference V(r c ) V(r b )? A. kq/r a B. 2kq/r a C. kq/r c D. 2kq[1/r a 1/r b ] E. zero F. 2kq[1/r a 1/r b ] G. 2kq[1/r a 1/r c ] H. 2kq[1/r a 1/r c ] 4) If the electric potential difference V(r) V(r a ) for r < r a is V(r) V(r a ) = 3 k q [r 2 /r a 2 1]/2 r a , what is the electric potential difference V(r) V( ) for r < r a ? A. 3kqr 2 /2r a 3 + 9kq/2r a B. 3kqr 2 /2r a 3 + 9kq/2r a 3kq/r b + kq/r c V(r c ) V(r b ) = b c E dl Since E = 0 in the conductor, V(r c ) V(r b ) = 0 For r < r a V(r) V( ) = [ V(r) V(r a )] + [V(r a ) V(r b )] + [V(r b ) V(r c )] + [V(r c ) V( )] Since E = 0 in the conductor, V(r b ) V(r c ) = 0 V(r) V(r a ) = 3 k q [r 2 /r a 2 1]/2 r a is given Electric field between r a and r b is the same as the electric field of a point charge + 3q; thus V(r a ) V(r b ) = 3 k q/r a 3 k q/r b Electric field for r > r c is the same as the electric field of a point charge q; thus V(r c ) V( ) = k q/r c Add up the contributions.
Background image of page 2
C. 3kqr 2 /2r a 3 + 9kq/2r a + kq/r c D. 3kqr 2 /r a 3 + 3kq/2r a 3kq/r b + kq/r c E. zero F. 3kqr 2 /r a 3 + 3kq/2r a G. 3kqr 2 /r a 3 + 3kq/2r a + kq/r c H. 3kqr 2 /r a 3 + 3kq/2r a 3kq/2r b + kq/r c 5) A positive charge Q is distributed in an insulating sphere. The radius of the sphere is r a . The magnitude of the electric field inside the sphere (r r a ) is E(r) = k Q r 2 /r a 4 . What is the electric potential difference V(r)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 15

2306_sol test 2 form A f07 - Physics 2306 Fall 2007 Second...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online