AP EAMCET · PHYSICS · Electrostatics
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
- A \(\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\)
- B \(V_A=V_B>V_C\)
- C \(\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} < \mathrm{V}_{\mathrm{C}}\)
- D \(V_A>V_B=V_C\)
Answer & Solution
Correct Answer
(B) \(V_A=V_B>V_C\)
Step-by-step Solution
Detailed explanation
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.…
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