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(B) In a uniform electric field,the electric potential decreases as we move in the direction of the electric field.Since $BC$ is parallel to the elect

Vedclass · Accepted Answer

$V_A=V_B>V_C$

Vedclass · Answer

(B) In a uniform electric field,the electric potential decreases as we move in the direction of the electric field.Since $BC$ is parallel to the electric field $\overrightarrow{E}$ and point $C$ is further along the direction of the field than point $B$,we have $V_B > V_C$.Also,the electric potential remains constant along a line perpendicular to the direction of the electric field.Since $AB$ is perpendicular to the electric field $\overrightarrow{E}$,the potential at $A$ is equal to the potential at $B$,i.e.,$V_A = V_B$.Combining these two results,we get $V_A = V_B > V_C$.

The figure shows three points $A, B$ and $C$ in a uniform electric field $(\overrightarrow{E})$. The line $AB$ is perpendicular to $BC$ and $BC$ is parallel to $\overrightarrow{E}$. If $V_A, V_B$ and $V_C$ are the potentials at $A, B$ and $C$ respectively,then the correct option is

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