Calculate the amount of work done in rotating a dipole, of dipole moment 3 x 10^-8cm,

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

The correct answer is (a) (d)  (a) The process must be adiabatic.  (d) The temperature must decrease.

The correct answer is (c) CA > CB

The correct answer is (d) Q = W

The correct answer is (a) (d)  (a) Q = 0.  (d) Q≠W.

At every point on equatorial axis, the potential is zero, so work done W = qΔV = 0(zero).

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)

Δ U = q – w, closed system