Van de Graaff generator is capable of producing very high potential of the order of 10 MV.

The underlying principle of Van de graaf generator is that a charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.



Construction: It consists of a large hollow metallic sphere S mounted on two insulating columns A and B and an endless belt of rubber or silk is made to run on two pulleys P₁ and P₂ by the means of an electric motor. C₁ and C₂ are two sharp metallic spikes in the form of combs. The lower comb C₁ is connected to the positive terminal of a very high voltage source and the upper comb C2 is connected to the inner surface of metallic sphere S.

Working: When comb C₁ is given very high potential, then it produces ions in its vicinity, due to action of sharp points. The positive ions, so produced, get sprayed on the belt due to the repulsion between positive ions and comb C₁. These positive ions are carried upward by the moving belt. The pointed end of C₂ just touches the belt. The comb C₂ collects positive charge from the belt which immediately moves to the outer surface of sphere S. As the belt goes on revolving, it continues to take (+) charge upward, which is collected by comb C₂ and transferred to outer surface of sphere S. Thus the outer surface of metallic sphere S gains positive charge continuously and its potential rises to a very high value.

When the potential of a metallic sphere gains very high value, the dielectric strength of surrounding air breaks down and its charge begins to leak, to the surrounding air. The maximum potential is reached when the rate of leakage of charge becomes equal to the rate of charge transferred to the sphere. To prevent leakage of charge from the sphere, the generator is completely enclosed in an earthed connected steel tank which is filled with air under high pressure.

Van de Graff generator is used to accelerate stream of charged particles to very high velocities. But, there is a restriction on the upper limit of high voltage set up in the machine. The high voltages can be built up to the breakdown field of the surrounding medium.

a)

The schematic sketch of cyclotron is as shown below:

 

Electric field: It helps in accelerating the charged particle passing through the gap with the help of an electric oscillator. An electric oscillator imparts the energy to charged particle till it comes out from the exit slit.

Magnetic field: The magnetic force exerts a centripetal force when the accelerated charge particle enters normally to the uniform magnetic field. Centripetal force makes the particle move in a semicircular path of increasing radii in each Dee.

Kinetic energy acquired by the particle is given by, 

 

b) 

i) Now, using equation (1), we have

 
Here we can see that, cyclotron frequency depends upon (q/m) ratio.

So, 

 

That is, 

 

ii) Kinetic energy is given by, 

 

That is, 



That is, proton acquires higher velocity.

When a conductor like a metal is subjected to external electric field, the electrons experience a force in the opposite direction and they accumulate on the left hand side. A positive charge is therefore induced on the right hand side. This creates on opposite electric field ( E_in ) that balances out (E₀). The net electric field inside the conductor becomes zero. 

Here, E₀ is the external field and E_in is the internal field created by the re-distribution of electrons inside the metal.

Dielctric: When external electric field is applied, dipoles are created for non-polar dielectrics and in case of polar dielectrics dipoles are aligned. An internal electric field is created such that it reduces the external electric field. The placement of dipoles is as shown in the given figure.

 

The dipole moment per unit volume of the polarized dielectric is known as the Polarization of a dielectric. It is denoted by P.

The formula for polarization is given by, 

b) i) The electric field inside a spherical shell is zero. Therefore, force on the charge placed at the centre of the shell is zero. For a charge placed at A, shell will behave as if the entire charge Q is placed at the centre of the shell.

Therefore, total charge is, 

Distance from 2Q is ‘x’. 


So, according to Gauss’s law, 

Total flux, 

Electric dipole moment is the product of either charges or the distance between two equal and opposite charges.

It is a vector quantity.

Electric dipole moment at a point on the equatorial plane:

Consider a point P on broad side on the position of dipole formed of charges + q and - q at separation 2l. The distance of point P from mid-point O of electric dipole is r. 

 

Let E₁ and E₂ be the electric field strength due to charges +q and –q of electric dipole.

From the fig. we have

 
Now, in order to find the resultant electric field, we resolve the components along and perpendicular to AB.

The components perpendicular to AB are sin components and they being equal and opposite to each other cancel each other.

Therefore,

Resultant electric field is given by, 

E₁ = E₁cos θ + E₂ cos θ

But, 

 

From the fig. we can see that,

  

If dipole is infinitesimal and point P is far away, then l² can be neglected as compared to r².

Therefore, 

 

b) Equipotential surfaces due to an electric dipole is given by,  
                               

Electric potential is zero at all points in the plane passing through the dipole equator.