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(B) The electric potential $V$ at any point inside or on the surface of a charged object is given by the integral of the potential contribution from a

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$V_A < V_O$

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(B) The electric potential $V$ at any point inside or on the surface of a charged object is given by the integral of the potential contribution from all surface charge elements $dq$,i.e.,$V = \int \frac{k dq}{r}$.For a hollow hemisphere with uniform surface charge density $\sigma$,the potential at the center $O$ is higher than at any other point on the base because the center is,on average,closer to the charged surface elements than any other point on the base.As we move from the center $O$ towards the rim $A$,the distance to the various parts of the charged surface increases on average,leading to a decrease in the electric potential.Therefore,the potential at the center $O$ is the maximum potential on the base of the hemisphere.Thus,$V_O > V_A$ or $V_A < V_O$.

Charge is uniformly distributed on the surface of a hollow hemisphere. Let $O$ and $A$ be two points on the base of the hemisphere,where $O$ is the center of the base and $A$ is a point between the center and the rim. Let $V_O$ and $V_A$ be the electric potentials at $O$ and $A$ respectively. Then,

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