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

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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

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$$1.8$$×$$106$$ $$=9$$×$$109$$×$$q1(0.1)2$$  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 $$isU=kq0.2($$−$$q)=9$$×$$109(2$$×$$10$$−$$6)20.2=$$−$$0.18$$  $$JSelf-energies$$ $$areU1=q28$$πε$$0r1=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.1=0.18$$  $$JU2=q28$$πε$$0r2=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.2=0.09$$  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 $$isU=kq0.2($$−$$q)=9$$×$$109(2$$×$$10$$−$$6)20.2=$$−$$0.18$$  $$JSelf-energies$$ $$areU1=q28$$πε$$0r1=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.1=0.18$$  $$JU2=q28$$πε$$0r2=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.2=0.09$$  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 $$isU=kq0.2($$−$$q)=9$$×$$109(2$$×$$10$$−$$6)20.2=$$−$$0.18$$  $$JSelf-energies$$ $$areU1=q28$$πε$$0r1=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.1=0.18$$  $$JU2=q28$$πε$$0r2=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.2=0.09$$  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 $$isU=kq0.2($$−$$q)=9$$×$$109(2$$×$$10$$−$$6)20.2=$$−$$0.18$$  $$JSelf-energies$$ $$areU1=q28$$πε$$0r1=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.1=0.18$$  $$JU2=q28$$πε$$0r2=9$$×$$109(2$$×$$10$$−$$6)22$$×$$0.2=0.09$$  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.

Electrostatic potential energy

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The electric potential at any point inside the first shell is

The electric field intensity just inside the outer sphere is

The electrostatic energy stored in the system is

If both the spheres are connected by a conducting wire, then

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