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Electric field due to an electric dipole on its - Consider, an electric dipole of length 2a having charges +q and  -q. A point P lies on the equatorial line of the dipole at a distance r from the centre of the dipole. The electric fields due to '-q' and '+q' are E1 and E2 respectively. Now, | E1 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(1) | E2 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(2) From equation (1) and (2) | E1 |  =  | E2 |  =   x (say) Clearly, components E1sinθ and E2sinθ cancel each other. Therefore, Net electric field at point P is - E = xCosθ + xCosθ = 2xCosθ From equation (1) E = 2kq.Cosθ/(r² + a²)         = kq2a/[(r² + a²).(r² + a²)½]   = kq2a/[(r² + a²)³]½ Also, dipole moment - Questions for Practice Q1. Find the ratio of electric fields due to a short dipole on the axial line to equatorial line at same distances from the centre. Ans.    Ea/Ee = (2kp/r³)/(kp/r³)                   = 2

Applications of Electrostatic shielding - 1. In a lightning thunderstorm, it is safe to sit inside a car rather than near a tree or in open ground. The metallic body of the car act as electrostatic shielding from the lightning to inside the car.  (Important point for boards) 2. In coaxial cable, the outer conductor connected to the ground provides an electrical shield to the signals carried by the central conductor. For electric potential due to an electric dipole CLICK HERE by    www.advguruji.blogspot.com

Consider, an uniformly charged infinite sheet having surface charge density σ . Let, A be the small area of the sheet containing charge q . As, charge lie on both the sides of the sheet, so electric field E is also on both the sides. using Gauss theorem, net electric flux through the cylinder enclosing the charge 'q' is -                       Φ    =  q/εο          ......(1) But,                Φ   =  Φ1 + Φ2 + Φ3                             = EACos0° + EACos0° + 0                        Φ  =  2EA      .......(2)           From equation (1) & (2)                 2EA = q/εο                  E    =   q/2Aεο  Also,          σ  = q/A  Therefore,                      [  E = σ/2εο  ] Still not clear click here to watch     by    www.advguruji.blogspot.com

Consider, an electric dipole having dipole moment p in an uniform electric field E such that the dipole moment is making an angle θ with the field. Now, small amount of work done dW in rotating the dipole by small angle dθ is - Total work done when dipole is rotated from angle θ1 to angle θ2 with the field is -                             =  -pE[Cosθ2 - Cosθ1]                        W =  pE[Cosθ1 - Cosθ2] This work done will be stored as potential energy of the dipole. Therefore,  U =  pE[Cosθ1 - Cosθ2] NOTE :- As we know, more the potential energy of a system, less it will be stable and vice versa. 1. For most stable condition    [U = minimum]        θ1 = 180° ,    θ2 = 0° Therefore, U = -2pE 2. For most unstable condition    [U = maximum]       θ1 = 0°  ,     θ2 = 180° Therefore, U = pE[1 - (-1)]                    U = 2pE Still not clear click here to watch      by  www.advguruji.blogspot.com

Statement : The net electric flux through any closed surface is equal to 1/εο times the charge enclosed by the surface.               Φ = q/εο where, q = charge enclosed by the surface PROOF consider, a positive point charge q which is enclosed by an imaginary spherical shell of radius r with 'q' at the centre. ds is a very very small area element of the shell. As we know, the electric field of a positive charge is always radially outward. Pyaare bacchon, agar online study se headache hota hai toh dekh lena, thoda aaram milega tumhe Now, small electric flux through 'ds' is - But, E = q/(4πεο.r²) Therefore, dΦ = [ q/(4πεο.r²) ].ds Now, Net electric flux through the whole spherical surface is - (Integrating both sides and taking limits) Question Find the value of x in the figure if the net flux through the Gaussian surface is '1/2εο' Nm²/C. For "electric field near an uniformly char

Best of Luck STUDENTS Que1. Two small conducting spheres of equal radius have charges 10μC & -20μC respectively placed at a distance R from each other. They experience force F1 . If they are brought in contact and separated to the same distance, they experience force F2 .   Ratio of F1 to F2 is ___ a) 1:2 b) 1:1 c) 2:1 d) 8:1 Pyaare bacchon, agar online study se headache hota hai toh dekh lena, aaram milega Que2. Electric field intensity at a point in between two parallel sheets with like charges of same surface charge densities σ is ___ a) σ/εo b) zero c) σ/2εo d) none of these Que3. There exists an electric field of 2 N/C along y direction. The flux passing through the square of side 1 m placed in xy plane inside the electric field is - a) Zero b) 2 Nm²/C c) 1 Nm²/C d) √3 Nm²/C Que4. A mass of 20g has a charge of 3mC . It moves with a velocity of 20m/s and enters a region of electric field of 80N/C in the same direction as the velocity of the mass. The velocity of t

Do   your best Q1. If a charged body A having charge +2q touched another uncharged body B . Then A touched another charged body C having charge +2q . Find the force between A and C if they are separated by 1 m. Q2. A system has two charges Q1 = 3μC and Q2 = -3μC . The coordinates of Q1 and Q2 are (10,4,2)m and (5,-9,-10)m respectively. (a) Name the system. (b) Find the net charge of the system. (c) Find the dipole moment of the system. Q3. find the force between the total negative and the total positive charge extracted from 100 ml of water and then kept at 10 cm apart. Q4. Find the nature and magnitude of  x  so that "P" is a null point. Q6. find the amount of two equal charges separated by 10 cm so that the force between them can support the weight of a 40 kg person. ( Take, g = 10 m/s² ) ---------------------------------------------------------------------------------------------- ANSWERS Ans1. , Ans2. , Ans3. (( all questions I have discu

Consider, a short electric dipole having dipole moment p . A point P is at a distance r from the centre O of the dipole such that the angle between p and OP is θ . Now, P act as a point on the axial line for pCosθ component and act as a point on the equatorial line for pSinθ component. The electric fields at point P due to pCosθ and pSinθ are E1 and E2 respectively. Also, they are perpendicular to each other. Now, net electric field at point P is - by www.advguruji.blogspot.com

Consider, an electric dipole of length 2a having charges +q and -q . A point P is lying on the axial line of the dipole at distance r from the centre of the dipole (as shown in figure). The electric fields due to charges -q and +q are E1 and E2 respectively. for Electric field due to an electric dipole on its equatorial line CLICK HERE by     www.advguruji.blogspot.com