solution: Let q be the charge on inner shell when it is earthed V_inner = 0.

i.e.  charge will flow from inner shell to earth.

Electric dipole: An arrangement of two equal and opposite charges separated by small distance is known as electric dipole.

                                                 

                                                            Fig (20)

1)      It is a vector quantity.
2)      It is directed from –ve charge to +ve charge.
3)      The line joining the charges is the axis of dipole.
a) Electric field on the axis (l<<r) =

     Derivation:

                                         

                                                            Fig (21)

 

 

b) On perpendicular bisector =

Derivation:

                                               

                                                         Fig (22)

    The component of E cancels out in vertical direction.
    Along horizontal.

    E_net = E_+qCosq + E_-qCosq

   
          (-ve indicates direction opposite to dipole moment)

         If x>> 1

          

 

 

c)      Electric field at any point A(r,0)

                                            

                                                  Fig (23) 

  

Dipole is an external uniform electric field
1)      Force on dipole=0.
2)     

3)  Potential energy = -PECos q = -P.E

                                    

                                                        Fig (24)

Q= 0, U will be minimum, stable equilibrium.
Q=180⁰, U will be maximum, unstable equilibrium.

Illustration:
1) A dipole of dipole moment P lies in uniform electric field with its dipole moment    along E. If dipole is rotated through 90⁰, Find work done?
U_i = -P.E = -P.E Cosq = -P.E.
U_f = -PECos90⁰ = 0
Work done = -DU = U_i-U_f =-P.E.

2) A dipole with dipole moment  Cm is aligned at 80⁰ with the direction of a electric field of magnitude 10⁴N/C Calculate magnitude of torque.

|Z| = PESinq

Examples:

1)   A charge q = 2´10^-6C is placed at (1m, 1m, 1m) and electric field at P (0m,-2m, 1m).
solution:

2)      Two free particles with charges q and 4q are a distance L apart. A third charge is placed so that system is in equilibrium. Find location, magnitude and size of third charge.
solution: Let new charge be +Q, It will be along the line joining the two charges to balance the forces. Let it be located as shown.

                         Fig (25)

Since they are in equilibrium
For q:

For 4q:

(-ve x means charge will be in middle of the two charges.)

3) A charged particle of mass m and charge q is released from rest in an electric field of constant magnitude E then the KE of the particle after time t is.

solution:

Force = qE

Acceleration = .

4) P and Q are two concentric metallic spheres. P is positively charged and Q is earthed then

a) Charge density on Q is same as on P.
b) Electric field between P and Q is uniform.
c) Electric potential inside P is zero.
d) Electric field outside Q is zero.

                                         

                                                    Fig (26)

solution: Let charge on P be q1 and on Q be q2
V_Q = 0

i.e. (a) is false.

(b) Is false as there is potential difference between the two shells and the electric field lines separate outwards.

(c) is false .

(d) is true as q enclosed by Q surface = q2+q1=0.

5) A certain charge Q is divided into two parts q and Q-q. For the maximum Coulomb force between them the ratio (q/Q) is:

1) 1/6, 2) 1/8, 3) 1/4, 4) 1/2

 

solution:

Ans = 4).

6) Three large conducting plate sheets are kept as shown in figure (27). Find electric field a) between (1) and (2), b) between (2) and (3).

                                                        

                                                           Fig (27)

solution:

a) Between (1) and (2)

Field due to a plate = .

b) Between (2) and (3)

7) A charged particle of mass m=2kg and q=1microcoulomb is thrown from ground at   q=30⁰ with speed 30 m/s. There is a horizontal electric field E=10⁷V/m exists. Find the range on horizontal ground of the projectile.
solution:

8) A charge q is placed at l/2 distance above a square of length l as shown in figure (28). Find flux through the square.

solution: Consider the charge to be kept in a cube of side l.

Net flux =
                                     

                                                     Fig (28)

 

 

By symmetry flux through each face = .

9) An alpha particle of KE = 6MeV is heading towards a stationary nucleus of atomic no 30 calculate distance of closet approach.

By conservation of energy:

½ mV²  = U_f-U_i

Dump Question: Why V_i = 0?
Ans: Since charges are at infinity