In the figure a solid sphere of radius a 3.10 cm is concentric with a spherical

Question

In the figure a solid sphere of radius a 3.10 cm is concentric with a spherical conducting shell of inner radius b 2.0 aand outerraau sc 2.4 a The sphere a ane un or charge q1 +5.08 fC; the shell has a net charge q2 =-q1, what is the magnitude of the electric field at radial distances (a) r-0 cm, (b): a/2.00, (c) r-a, (d) r-1.50a, (e) r = 2.3 and (f) r3.50a? What is the net charge on the (9) inner and (h) outer surface of the shell? (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (9) Number (h) Number Units Units Units Units Units Units Units

Explanation / Answer

Please Concentrate.

A)We need to find the magnitude of electric field at r=0 hence r<a hence the formula we need to apply is for the E=q/(4*pi*epsilon0*a3)*r and that will be for b as well as in that case as well r<a.For the part c we will use the formula  E=q/(4*pi*epsilon0*a2 ) as r= a and for the rest of the parts if required the formula to be used will be E=q/(4*pi*epsilon0*R2) as r>=a

Using the formula as mentioned above we get .

a) E=0 as r= 0 cm.

b) E=5.62*10-2N/C

c) E=0.112N/C

d)E=0.05 N/C

e) E=0 N/C Why ?big question.Now concentrate, we know that b= 2.00a and c= 2.40a hence the point should lie within the radius b and c as for part d r=2.30a.And by gauss law E inside a conductor is always equal to zero.

f) E=0 N/C as the total charge outside the spherical shell is -q hence Q=+q +(-q)= zero putting that into the above equation we get E=0.

As are subjected to 2 decimal places internal calculation.

B) We know that the Electric field inside a conductor is 0 hence the electric flux is also 0 hence the charge enclosed is also 0 by gauss law.

Hence the Qsphere+Qinner=0

Therefore the Qinner  =- Qsphere = - 5* 10-15 C (ANS g)

now (Qinner shell+Qoutershell=- 5 FC)

therefore the Qoutershell = 0 (ANS h)

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