A solid insulating sphere of radius a carries a net positive charge `3Q`, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c and having a net charge -Q, as shown in figure

a. Consider a spherical Gaussian surface of radius rgtc, the net charge enclosed by this surface is .............
b. The direction of the electric field rgtc is .............
c. The electric field at rgtc is ............... .
d. The electric field in the region with radius r, which cgtrgtb, is ...............
e. Consider a spherical Gaussian surface of radius r, where `cgtrgtb`, the net charge enclosed by this surface is ................ .
f. Consider a spherical Gaussian surface of radius r, where`bgtrgta`, the net charge enclosed by this surface is ................. .
g. The electric field in the region `bgtrgta` is ................ .
h. Consider a spherical Gaussian surface of radius `rlta`. Find an expression for the net charge Q(r) enclosed by this surface as a function of r. Note that the charge inside the surface is less than 3Q.
i. The electric field in the region `rlta` is ................. .
j. The charge on the inner surface of the conducting shell is .......... .
k. The charge on the outer surface of the conducting shell is .............. .
l. Make a plot of the magnitude of the electric field versus r.

Explanation

The distribution of charge is as shown in figure.

Net charge enclosed by Gaussian surface `(rgtc)` is 3Q
`-3Q+2Q = 2Q.`
b. For rgtc, the direction of electric field is radially
outward as shown at point A.
c. `E = (kq_(in))/(r^2)= 1/(4piepsilon_0) (2Q)/(r^2)`
d. For cltrltb, we have point B. If we draw a Gaussian
surface through B, net charge enclosed in it will be zero.
So electric field at B will be zero.
e. As explained above, net charge enclosed is zero.
f. For bgtrgta, we have point D. If we draw a Gaussian
surface through D, it will have net charge enclosed 3Q.
g. `E = (kq_(in))/(r^2) = 1/(4piepsilon_0) (3Q)/(r^2)`
h. `rho = (3Q)/(4/3pia^3)` (volume charge density)
So, charge with in radius `r(lta)` is `q = rho4/3 pir^3 = (3Qr^3)/a^3`
i. `E = (kq_(in))/r^2 = 1/(4piepsilon_0)(3Qr^3)/(a^3r^2) = 1/(4piepsilon_0)(3Qr)/(a^3)`
j. As shown in the figure, the charge on the inner surface
of shell is `-3Q`.
k. As shown in figure, the charge on outer surface of
shell is +2Q.
l. .