What is the work done in moving a test charge q through a distance of 1cm along the equatorial axis of an electric dipole?

Electric dipole moment, p = 4 × 10−9 C m Angle made by p with a uniform electric field, θ = 30° Electric field, E = 5 × 104 N C−1 Torque acting...

this is formula based question..... since we know,  v= w/q therefore, electric potential (v)=12.5/75 v=125/75o    =1/6   =0.16  ans

Work done = 10 Joule Charge = 2 Coulomb Potential of point A = VA = 10J/2C Given, work done = 20 Joule=5V Charge = 2 Coulomb Potential of point B = VB =20J/2C=10V Potential difference...

The correct answer is (a) zero

The correct answer is (b) (c) (b) may be equal to 20V/cm (c) may be greater than 20V/cm

(d) The torque on the dipole due to the field may be zero.

The elecrostatic potential due to an electric dipole at an equatorial point is Zero.

P = 3 x 10-8cm ; E =104N/C  At stable equilibrium (θ1 ) = 0°  At unstable equilibrium (θ2) =180°  Work done in rotating dipole is given by:  W = PE (cosθ1 - cosθ2 )   =...

Dipole moment P =5 x10-4Cm. Electric field (E) =104N/C At stable equilibrium θ1 = 0° At unstable equilibrium θ2 =180° Work done = - PE (cosθ2 - cosθ1) = -(5 x 10-8 ) (104 )...

Dipole moment of an electric dipole is the product of either of charge and the length of dipole. It is a vector quantity. vector P = q(2vector a)