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

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(B) $1$. In a uniform electric field,the potential does not change along a direction perpendicular to the field lines. Since $AB$ is perpendicular to

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$V_A = V_B > V_C$

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(B) $1$. In a uniform electric field,the potential does not change along a direction perpendicular to the field lines. Since $AB$ is perpendicular to the electric field lines,the potential at $A$ is equal to the potential at $B$,i.e.,$V_A = V_B$.$2$. The electric potential decreases in the direction of the electric field. Since $BC$ is parallel to the electric field lines and points in the direction of the field,the potential at $C$ is less than the potential at $B$,i.e.,$V_B > V_C$.$3$. Combining these two results,we get $V_A = V_B > V_C$.

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