Work done depends on the angle of rotation and the strength of the electric field.
Step 1: Formula for work done - Work done is given by: \[ W_{\text{ext}} = U_f - U_i, \] where \(U = -\vec{P} \cdot \vec{E}\). - Initial potential energy: \[ U_i = -PE \cos 0 = -PE. \] - Final potential energy: \[ U_f = -PE \cos 180 = +PE. \]
Step 2: Substitute the values - \[ W_{\text{ext}} = U_f - U_i = PE - (-PE) = 2PE. \] Substituting values: \[ W_{\text{ext}} = 2 \cdot 6.0 \times 10^{-6} \cdot 1.5 \times 10^3. \] Simplifying: \[ W_{\text{ext}} = 18 \, \text{mJ}. \]
Final Answer: The work done is 18 mJ.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC? 
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)