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.
An infinite sheet of uniform charge \( \rho_s = 10\, {C/m}^2 \) is placed on the \( z = 0 \) plane. The medium surrounding the sheet has a relative permittivity of 10 . The electric flux density, in C/m\(^2\), at a point \( P(0, 0, 5) \), is: Note: \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors along the \( x, y, z \) directions, respectively.
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?
Match List-I with List-II: List-I