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 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? 
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: