Gauss law An insulating sphere of radius a has a total charge -3Q uniformly dist

Question

Gauss law

An insulating sphere of radius a has a total charge -3Q uniformly distributed over its volume, as shown below. The sphere is surrounded by a conducting spherical shell of inner radius 2a and outer radius 3 a. The total charge on the two surfaces of the shell is unknown. a) Using Gauss' Law explicitly (not simply plugging variables into a result derived in class), find the magnitude and direction of the electric field at a distance r = a/9 from the center of the sphere (in region I). Express your answer in terms of Q, a, and any other relevant constants. b) As one moves from the center of the insulating sphere (at r = 0) to its boundary (at r = a), does the value of electric potential increase, decrease, or remain constant? Justify your answer briefly. b) The electric field at a distance r = 4a from the center of the sphere (in region IV) has magnitude 1/4 pi epsilon_0 Q/6a^2 = k_e Q/6a^2 and is directed radially outward. Again, using Gauss' Law, determine the charge on the outer surface of the conducting shell. State your answer in terms of Q.

Explanation / Answer

given,

insulating sphere of radius a, charge -3Q, uniformly distributed over its volume

a) consider a gaussean surface, spherical at radius r = a/9 from the center

now volume enclosed inside the gaussean surface, V = 4*pi*r^3/3 = 4*pi*a^3/3*729

total volume of the insulating sphere, V' = 4pi*a^3/3

so net charge enclosed inside the gausean surface, qin = -3Q*V/V' = -3Q(4*pi*a^3/3*729 ) / 4pi*a^3/3 = -3Q/729 = -Q/243

le tthe electric field at this point be E

then from gauss' law

E*4*pi*r^2 = qin/epsilon

E = 27*qin/epsilon*4*pi*a^2 = 27*(-Q)/4*pi*epsilon*a^2*243 = -kQ/9a^2 [ here k = 1/4*pi*epsilon , coloumb's constant]

b) at raidus r < a

qin = -3Q*4*pi*r^3*3/3*4*pi*a^3 =-3Qr^3/a^3

so E = -3Qr^3/a^3*4*pi*r^2*epsilon = -3Qr/a^3*4*pi*epsilon = -3Qkr/a^3

so as r increases, E becomes more -ve

now E = -dV/dr

V = - integral Edr = - integral -3Qkr dr/a^3 = 3Qk[r^2]/a^3

so as r increases, Potential also increases

c) if charge on the outer surface of the conducting shell is q

then on the inner surface the charge is 3Q [ as electric field inside conductor is 0, so charge on inner surface of conducting shell and that on the insulating sphere inside must be 0]

so, from gauss' law

E*4*pi*(4a)^2 = q/epsilon

E = q/4*pi*epsilon*16a^2 = Q/4*pi*epsilon*6a^2

q/16 = Q/6

q = 16Q/6

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