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Figure shows a charged conductor resting on an insulating stand. If at the point P the charge density is \sigma, the potential is V and the electric field strength is E, what are the values of these quantities at point Q

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As we learned

Surface charge distribution -

\left ( \sigma \right )- charge per unit Area 

\dpi{100} \sigma=\frac{Q}{A}=\frac{C}{m^{2}}=Cm^{-2}

 

- wherein

(Plane sheet, sphere, cylinder etc)

 

 The surface of the conductor is an equipotential surface since there is free flow of electrons within the conductor. Thus potential at Q is the same as that at P. That is V_{P}=V_{Q}=V.  The electric field E at a point on the equipotential surface of the conductor is inversely proportional to the square of the radius of curvature r at that point. That is E\, \alpha \, r^{-2}   Since point Q has a larger radius of curvature than that at point P, the electric field at Q is less than that at P. That is E_{Q}<E_{P}=E  

 


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