A conducting sphere of radius R is placed in uniform electric field $${\overrightarrow {\bf{E}} _0}$$ directed along +Z-axis. The electric potential for outside points is given as $${V_{{\text{out}}}} = - {E_0}\left( {1 - \frac{{{R^3}}}{{{r^3}}}} \right)r\cos \theta ,$$      where r is the distance from the centre and θ is the polar angle. The charge density on the surface of the sphere is

A conducting sphere of radius R is placed in uniform electric field $${\overrightarrow {\bf{E}} _0}$$ directed along +Z-axis. The electric potential for outside points is given as $${V_{{\text{out}}}} = - {E_0}\left( {1 - \frac{{{R^3}}}{{{r^3}}}} \right)r\cos \theta ,$$      where r is the distance from the centre and θ is the polar angle. The charge density on the surface of the sphere is Correct Answer 3ε<sub>0</sub> E<sub>0</sub> cos θ

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A rod of length L with uniform charge density $$\lambda $$ per unit length is in the XY-plane and rotating about Z-axis passing through one of its edge with an angularvelocity $$\overrightarrow \omega $$ as shown in the figure below. $$\left( {{\bf{\hat r}},\,\hat \phi ,\,{\bf{\hat z}}} \right)$$   refer to the unit vectors at Q, $$\overrightarrow {\bf{A}} $$ is the vector potential at a distance d from the origin O along Z-axis for d ≪ L and $$\overrightarrow {\bf{J}} $$ is the current density due to the motion of the rod. Which one of the following statements is correct?
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