The figure shows three points A, B and C in a | vedclass

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(B) According to the given figure,the line $A B$ is perpendicular to the direction of the electric field lines. Hence,the surface passing through line

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$V_{A}=V_{B}>V_{C}$

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(B) According to the given figure,the line $A B$ is perpendicular to the direction of the electric field lines. Hence,the surface passing through line $A B$ and perpendicular to the electric field lines behaves like an equipotential surface,therefore$V_{A}=V_{B} \dots (i)$Electric field and electric potential are related as$E=-\frac{d V}{d x} \Rightarrow V=-\int E d x$This indicates that the electric potential decreases in the direction of the electric field,i.e.$V_{B}>V_{C} \dots (ii)$From Eqs. $(i)$ and $(ii)$,we have$V_{A}=V_{B}>V_{C}$

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