In the figure a solid sphere of radius a = 2.20 cm is concentric with a spherica

Question

In the figure a solid sphere of radius a = 2.20 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q1 = +4.92 fc; the shell has a net charge q--ql, what is the magnitude of the electric field at radial distances (a) r = 0 cm, (b) r = a/2.00, (c) r = a, (d) r = 1.50a, (e 2.30a, and (f) 3.50a? What is the net charge on the (g) inner and (h) outer surface of the shell? (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (g) Number (h) Number Units Units Units Units Units Units Units Units

Explanation / Answer

E=(kq/a3)r uniform charge, field at r a

where a is the radius of solid sphere and k=8.98*10^9 Nm2/C2

GIven q=4.92*10-15C

a)r=0 cm

E=(kq/a3)r =(kq/a3)0=0

b) r=a/2

E=(kq/a3)r

= (8.98*10^9*4.92*10-15/a3)a/2 given r=a/2

substitute a=2*10-2 m

we get E=5.5227*10-2 N/C

c) r=a

E=(kq/a3)r =[ (8.98*10^9*4.92*10-15/a3)]a

sunstitute a=2*10-2 m

we get E= 0.11N/C

d) E=(kq/r2) (spherical, field at r a)

given r=1.5a

= 8.98*10^9*4.92*10^-15/1.5*(2*10^-2)^2

E= 0.04909 N/C

e) As per the given vales the point is located in b<r<c

Electric field inside the conducting shell is zero.

f) As per the given vales the point is located completely outside the figure.

Here sum of the charges is zero because q2=- q1

E=(kq/r2) here sum of charges is zero.

E=0

g) Net charge on the inner surface of the shell

E inside the shell is zero and so that flux is zero.

Gauss law E*flux=q ext which is equal to zero

qsphere+qinner=0

qinner = -4.92*10-15C

h) total charge of the shell = -4.92fC

qi+qo = -4.92fC

-4.92fC+qo=-4.92fC

qo = 0C

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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