14. +0/4 points | Previous Answers SerPSE9 24.P.057.WI My Notes Ask Your Teacher

Question

14. +0/4 points | Previous Answers SerPSE9 24.P.057.WI My Notes Ask Your Teacher For the configuration shown in the figure below, suppose a = 5.00 cm, b 20.0 cm, and c = 25.0 cm. Furthermore suppose the electric field at a point 11.5 cm from the center is measured to be 3.50 x 103 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 196 N/C and points radially outward. From this information, find the following. (Include the sign of the charge in your answer.) Insulator Conductor (a) the charge on the insulating sphere 0 Do the charges on the conducting shell affect the electric field at this point? C (b) the net charge on the hollow conducting sphere 0 What is the total charge inside a sphere of radius 50 cm? C (c) the charge on the inner surface of the hollow conducting sphere (d) the charge on the outer surface of the hollow conducting sphere Need Help? ReadIt Watch It Submit AnswerSave Progress Practice Another Version

Explanation / Answer

a)

From Gauss law

E=KQ/r2

3.5*103=(9*109)*Qinsulator/0.1152

Qinsulator=-5.14*10-9 C

where negative sign indicates direction of electric field

b)

From Gauss law

E=KQtotal/r2

196=(9*109)*Qtotal/0.52

QTotal=5.44*10-9 C

Charge on the conductor is

QConductor =Qtotal-Qinsulator=(5.44*10-9)-(-5.14*10-9)

QConductor =10.58*10-9 C

c)

Charge on inner surface

Qinner=Qinsulator=5.14*10-9 C

d)

Charge on outer surface

Qouter=Qconductor-Qinner=(10.58*10-9)-(5.14*10-9)

Qouter=5.44*10-9 C

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