The figure shows three points $A, B$ and $C$ in a uniform electric field $(\overrightarrow{\mathrm{E}})$.…

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

Solution

On moving in direction of electric field, electric potential decreases. So, $\mathrm{V}_{\mathrm{B}}>\mathrm{V}_{\mathrm{C}}$ Also in direction moving perpendicular to the direction of electric field. There is no change in potential. $\begin{aligned} & V_B=V_A \\ & V_A=V_B>V_C\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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