We have an isolated conducting spherical shell of radius 10 cm. Some positive charge is given to it so that the resulting electric field has a maximum intensity of 1.8×106  NC−1. The same amount of negative charge is given to another isolated conducting spherical shell of radius 20 cm. Now, the first shell is placed inside the second so that both are concentric as shown in figure.The electric potential at any point inside the first shell isThe electric field intensity just inside the outer sphere isThe electrostatic energy stored in the system isIf both the spheres are connected by a conducting wire, then

1.8×106 =9×109×q1(0.1)2  or  q1=2×10−6  CV=9×109×2×10−6[10.1−10.2]=9×104  VE=kq1r2=9×109×2×10−6(0.20)2=4.5×105  NC−1Interaction energy isU=kq0.2(−q)=9×109(2×10−6)20.2=−0.18  JSelf-energies areU1=q28πε0r1=9×109(2×10−6)22×0.1=0.18  JU2=q28πε0r2=9×109(2×10−6)22×0.2=0.09  JTotal energy is U+U1+U2=0.09  JBoth the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.1.8×106 =9×109×q1(0.1)2  or  q1=2×10−6  CV=9×109×2×10−6[10.1−10.2]=9×104  VE=kq1r2=9×109×2×10−6(0.20)2=4.5×105  NC−1Interaction energy isU=kq0.2(−q)=9×109(2×10−6)20.2=−0.18  JSelf-energies areU1=q28πε0r1=9×109(2×10−6)22×0.1=0.18  JU2=q28πε0r2=9×109(2×10−6)22×0.2=0.09  JTotal energy is U+U1+U2=0.09  JBoth the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.1.8×106 =9×109×q1(0.1)2  or  q1=2×10−6  CV=9×109×2×10−6[10.1−10.2]=9×104  VE=kq1r2=9×109×2×10−6(0.20)2=4.5×105  NC−1Interaction energy isU=kq0.2(−q)=9×109(2×10−6)20.2=−0.18  JSelf-energies areU1=q28πε0r1=9×109(2×10−6)22×0.1=0.18  JU2=q28πε0r2=9×109(2×10−6)22×0.2=0.09  JTotal energy is U+U1+U2=0.09  JBoth the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.1.8×106 =9×109×q1(0.1)2  or  q1=2×10−6  CV=9×109×2×10−6[10.1−10.2]=9×104  VE=kq1r2=9×109×2×10−6(0.20)2=4.5×105  NC−1Interaction energy isU=kq0.2(−q)=9×109(2×10−6)20.2=−0.18  JSelf-energies areU1=q28πε0r1=9×109(2×10−6)22×0.1=0.18  JU2=q28πε0r2=9×109(2×10−6)22×0.2=0.09  JTotal energy is U+U1+U2=0.09  JBoth the charges will get neutralized and there will be no charge left on any sphere. So no energy will be left in the system. It means whole amount of energy will convert into heat.