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.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32