Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

\"https://s3.amazonaws.com/answer-board-image/42a0f06c-f7e5-4365-97da-ad86567f02

ID: 2111068 • Letter: #

Question

"https://s3.amazonaws.com/answer-board-image/42a0f06c-f7e5-4365-97da-ad86567f0245.jpeg"

alt="uploaded image" border="0" />

Two concentric spheres are shown in the figure. The inner sphere

is a solid nonconductor and carries a charge of +5.00

µC uniformly distributed over its outer surface. The

outer sphere is a conducting shell that carries a net charge of

-8.00 µC. No other charges are present. The radii

shown in the figure have the values

src="http://session.masteringphysics.com/render?var=R_1" alt="R_1"

align="middle" /> = 10.0 cm,

"R_2" src="http://session.masteringphysics.com/render?var=R_2" alt=

"R_2" align="middle" /> = 20.0 cm,

"http://session.masteringphysics.com/render?var=R_3" alt="R_3"

align="middle" /> = 30.0 cm.

PART A: Find the total excess charge on the inner

and outer surfaces of the conducting sphere.

PART B: Find the magnitude and direction of the electric field at

"http://session.masteringphysics.com/render?var=r" alt="r" align=

"middle" /> = 9.5 cm from the center of the inner

sphere.

PART C: Find the magnitude and direction of the electric field at

"http://session.masteringphysics.com/render?var=r" alt="r" align=

"middle" /> = 15.0 cm from the center of the inner

sphere.


PART C: Find the magnitude and direction of the electric field at

"http://session.masteringphysics.com/render?var=r" alt="r" align=

"middle" /> = 35.0 cm from the center of the inner

sphere.

Explanation / Answer

I can't see the picture, sorry