1. A insulating solid sphere of radius 5 cm carries a uniform charge density giv

Question

1. A insulating solid sphere of radius 5 cm carries a uniform charge density given as p ( r) = p0r^2 where p0 = 5.0 x 10^2 C/m^5 and r is the radial distance measured from the center of the sphere. This insulating sphere is placed at the center of a concentric hollow conducting sphere (inner radius = 15 cm and outer radius = 25 cm) that carries an excess charge. Q = - 8.0 x 10^-8 C. By constructing an appropriate Gaussian surface. apply Gauss s law to determine the magnitude and direction of the electric field at the radial point. r = 30 cm from the common center of spheres.

Explanation / Answer

The charge on the insulating solid sphere is given by;

r(r)=r0 r2

r(r)=(5*10-2 C/m2)* (5*10-2m)2

r(r)=125*10-6 C

As this insulating sphere is placed at the center of hollow conducting sphere (inner radius 15 cm and outer radius 25 cm), that carries an excess charge of 8*10-8 C.

Electric flux in Gaussian surface assumed to be a sphere with radius 30 cm is given by;

f=Q/?0

f=8*10-8 C/8.85*10-12 F/m        (AS ?0=8.85*10-12 F/m)

f=0.9039*1012 C.m/F

As f=E.S

So,

E=f/S

E=0.9039*1012 C.m/F/(S)

Here; S=surface area of Gaussian surface=4pR2=4*3.14*(30 *10-2 m)2=11304*10-4 m2

SO,

E=0.9039*1012 C.m/F/S

E=(0.9039*1012 C.m/F)/(11304*10-4 m2)

E=7.99*1011 N/C

Now electric field due to insulating sphere can be found as;

dE

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