|
solution: Let q be the charge on inner shell when it is earthed Vinner = 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.
Enet = 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?
Ui = -P.E = -P.E Cosq = -P.E.
Uf = -PECos900 = 0
Work done = -DU = Ui-Uf =-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
VQ = 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 = Uf-Ui
Dump Question: Why V
i
= 0?
Ans: Since charges are at infinity
|
