21. 014 points I Previous Answers SarPSET9 24 P067 For the configuration shown i

Question

21. 014 points I Previous Answers SarPSET9 24 P067 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 My Notes Ask Your 0 cm, and c 25.0 cm. Furthermore, suppose the 17.0 cm from the center is measured to be 3.70 × 103 N/C radially inward and the electric field at a point 50.0 cm from the center is of magnitude 240 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 Do the charges on the conducting shell affect the electric field at this point? C (b) the net charge on the hollow conducting sphere (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? Read Watch t Subrnit Ansaver Save ProgressPractie Save Progress Practice Another Version Submit Save Home My Assignments Extension Request

Explanation / Answer

part a:

using gauss' law:

total flux passing through a surface=total charge enclosed by the surface

consider a gaussian surface of a sphere of radius 17 cm

if D=electric flux density

then total flux crossing this gaussian surface=D*4*pi*0.17^2=0.3632*D

as total charge enclosed by a sphere of radius 17 cm=total charge on the insulating sphere

==>charge on the insulating sphere=0.3632*D

now given that electric field at a point 17 cm from the center =3.7*10^3

we know that D=epsilon*E where epsilon=electrical permitivity of the medium

==>D=8.85*10^(-12)*3.7*10^3=3.2745*10^(-8)

hence magnitude of charge on the insulating sphere=0.3632*3.2745*10^(-8)=1.1892*10^(-8) C

as the electric field is directed radially inward, the charge has to be negative

hence total charge on the insulator=-1.1892*10^(-8) C

part b:

using the same method employed in part a,

charge enclosed=D*4*pi*0.5^2=3.1416*D

D=epsilon*E=2.124*0^(-9)

==>charge enclosed by the gaussian sphere=3.1416*D=6.67275*10^(-9) C

as the electric field is pointed radially outwards, the total charge enclosed by the gsaussian sphere of radius 50 cm

is positive

as total charge enclosed by this gaussian sphere=charge on the insulating sphere+charge on the holow conductor

==>6.67275*10^(-9)=charge on hollow conductor-1.1892*10^(-8)

==>charge on hollow conductor=1.8565*10^(-8) C

part c:

as electric field inside a conductor is zero,

considering a gaussian surface of radius =inner radius of hollow conductor

total flux density=0

==>total charge enclosed=0

==>charge on inner surface of hollow conducting sphere+charge on insulating sphere=0

==>charge on inner surface =-charge on insulating sphere=1.1892*10^(-8) C

part d:

as charge on hollow conducting sphere=charge on inner surface+charge on outer surface

==>charge on outer surface=charge on hollow conducting sphere-charge on inner surface

=1.8565*10^(-8)-1.1892*10^(-8)=6.673*10^(-9) C

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.