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Electrostatic potential energy

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Question

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  NC1. 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.

Moderate
Question

The electric potential at any point inside the first shell is

Solution

1.8×106=9×109×q1(0.1)2  or  q1=2×106  C

V=9×109×2×106[10.110.2]=9×104  V

E=kq1r2=9×109×2×106(0.20)2=4.5×105  NC1

Interaction energy is

U=kq0.2(q)=9×109(2×106)20.2=0.18  J

Self-energies are

U1=q28πε0r1=9×109(2×106)22×0.1=0.18  J

U2=q28πε0r2=9×109(2×106)22×0.2=0.09  J

Total energy is U+U1+U2=0.09  J

Both 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.

Question

The electric field intensity just inside the outer sphere is

Solution

1.8×106=9×109×q1(0.1)2  or  q1=2×106  C

V=9×109×2×106[10.110.2]=9×104  V

E=kq1r2=9×109×2×106(0.20)2=4.5×105  NC1

Interaction energy is

U=kq0.2(q)=9×109(2×106)20.2=0.18  J

Self-energies are

U1=q28πε0r1=9×109(2×106)22×0.1=0.18  J

U2=q28πε0r2=9×109(2×106)22×0.2=0.09  J

Total energy is U+U1+U2=0.09  J

Both 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.

Question

The electrostatic energy stored in the system is

Solution

1.8×106=9×109×q1(0.1)2  or  q1=2×106  C

V=9×109×2×106[10.110.2]=9×104  V

E=kq1r2=9×109×2×106(0.20)2=4.5×105  NC1

Interaction energy is

U=kq0.2(q)=9×109(2×106)20.2=0.18  J

Self-energies are

U1=q28πε0r1=9×109(2×106)22×0.1=0.18  J

U2=q28πε0r2=9×109(2×106)22×0.2=0.09  J

Total energy is U+U1+U2=0.09  J

Both 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.

Question

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

Solution

1.8×106=9×109×q1(0.1)2  or  q1=2×106  C

V=9×109×2×106[10.110.2]=9×104  V

E=kq1r2=9×109×2×106(0.20)2=4.5×105  NC1

Interaction energy is

U=kq0.2(q)=9×109(2×106)20.2=0.18  J

Self-energies are

U1=q28πε0r1=9×109(2×106)22×0.1=0.18  J

U2=q28πε0r2=9×109(2×106)22×0.2=0.09  J

Total energy is U+U1+U2=0.09  J

Both 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.


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Similar Questions

Two point charges +Q each have been placed at the positions (-a/2,0, 0) and (a/2,0, 0). The locus of the points where -Q charge can be placed such that total electrostatic potential energy of the system can become equal to zero, can be represented by which of the following equations?


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