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.
Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$ The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :
Consider the following data for the given reaction
\(2\)\(\text{HI}_{(g)}\) \(\rightarrow\) \(\text{H}_2{(g)}\)$ + $\(\text{I}_2{(g)}\)
The order of the reaction is __________.