IMPORTANT QUESTIONS OF ELECTROSTATICS

                        IMPORTANT QUESTIONS

Q.1-  Give two properties of electric lines of force sketch them for an isolated positive charge ?

Q.2- Derive an expression for dipole field intensity at any point on – (a)- axial line of dipole

         (b)- equatorial line of dipole.

Q.3-Derive an expression for torque acting  on  an electric dipole in a uniform electric field.

Q.4-Define capacitance and write its S.I. unit .

Q.5- Define an expression for the capacitance  of a parallel plate capacitor.

Q.6- Prove that energy density stored in parallel plate capacitor is1/2 є₀E²  where symbols have their usual meaning.

Q.7-  Derive an expression for energy stored in parallel plate capacitor . What is the form of this energy & where from it comes .

Q.8-   Find an expression for capacitance of parallel plate capacitor when dielectric slab is introduced between the plates of the capacitor .

Q.9-  It states gauss’s law in electrostatics. Using this law derive an expression for electric field due to charged spherical shell.

Q.10- Draw the level diagram of Van de graff generator . Explain the principle, working of  Van de graff generator.

VAN DE GRAFF GENERATOR

It was designed by Vande graff in the year 1931

PRINCIPLE-  The generator is based on –

1-       The action of sharp points  i.e,  phenomenon of CORONA DISCHARGE

2-       The property that charge is given to a hollow conductor is transferred to outer surface and distributed informally over it.

It is machine capable of building up a potential difference of few million volts & field close to the break down field of air which is about 3×10⁸ V/m

                                   A large spherical conducting shell (few meters radius ) is supported at height several meters above the ground on the insulating column.

A long narrow endless belt insulating material like rubber or silk is wound around two pulleys . One at ground level , one at the centre of the shell. This belt is kept continuously  moving by motor driving the lower pulley . It continuously carries positive sprayed on to it by a brush at ground level to the top up. There it transfers its positive charge to another conducting brush connected to the large shell. Thus the positive charge is transferred to the shell where it spread out on the outer surface.

Application- Accelerated positive charge are used to carried out nuclear reactions.

APPLICATION OF GAUSS LAW

CAPACITANCE OF A PARALLEL PLATE CAPACITOR WHEN DIELECTRIC SLAB IS INTRODUCED BETWEEN THE SLAB

LET ,E₀ is applied electric field

                        E₀  =  q/ є₀A     -----------------------(1)

                        E = Induced electric field

             Potential difference across the capacitor

                      V = E  ×t + E₀ (d-t)

            From definition of dielectric constant

                     K = E₀ / E

                Or  E = E₀/K

            On substituting the value of E

                V =   E₀t/K +  E₀ (d-t)

                 V   = E₀ [t/k + (d-t)]              -------------------(2)

          As     c= q/V             ----------------------(3)

           Now from equation (1)  in (2)

                 V = q/ є₀A  [t/k + (d-t)]            ----------------(4)            

Now, Equation (4) in (3)

               c = q/ (q/ є₀A  [t/k + (d-t)])           

                c = є₀A/[t/k + (d-t)]           

 If  t = d

       So  c = k є₀A/d

When battery is removed & dielectric slab is introduced the charge will be same.

APPLICATION OF GAUSS LAW

ENERGY DENSITY STORED BETWEEN THE PLATE OF THE CAPACITOR

Suppose the conductor is charged gradually at any stage , the charge on the conductor is q

Potential difference (v) = q/c

Small amount of work done in given an additional charge dq to the conductor

                       dw = q/c × dq

Total work done in giving a charge Q to the conductor  is under limit q=Q,q=0

                        W =  ∫ q/c dq  = 1/c [q²/2]^q=Q_q=0

                         W = 1/c × Q²/2

 Energy stored in the conductor

                       U = W = 1/c × Q²/2

                    As    Q = cV

                    So   U = 1/2 × (cV) ²/c = 1/2 cV²

                                   cV = Q ,            W = ½ QV

Energy density is defined as total energy stored per unit volume of the conductor

            u = total energy / volume

               = 1/2 cV²/Ad

            c = є₀A/d  and  V = E . d

               =  ½  (є₀A/d  ) (E² d²/Ad)

       Energy density = 1/2 є₀ E²