KCET · Physics · Electrostatics
Figure shows three points \(A, B\) and \(C\) in a region of uniform electric field \(\mathbf{E}\). The line \(A B\) is perpendicular and \(B C\) is parallel to the field lines. Then, which of the following holds good ? \(\left(V_{A}, V_{B}\right.\) and \(V_{C}\) represent the electric potential at points \(A, B\) and \(C\), respectively)
- A \(V_{A}=V_{B}=V_{C}\)
- B \(V_{A}=V_{B}>V_{C}\)
- C \(V_{A}=V_{B} < V_{C}\)
- D \(V_{A}>V_{B}=V_{C}\)
Answer & Solution
Correct Answer
(B) \(V_{A}=V_{B}>V_{C}\)
Step-by-step Solution
Detailed explanation
According to given figure, line \(A B\) is perpendicular to direction of electric field lines. Hence, surface passing through line \(A B\) and perpendicular to electric field lines behaves like a equipotential surface, therefore
\(V_{A}=V_{B}...(i)\)
Electric field and electric potential are related as
\(E=-\frac{d V}{d x} \Rightarrow V=-\int E d x\)
Which indicates that electric potential decreases in the direction of electric field, i.e.
\(V_{B}>V_{C}...(ii)\)
From Eqs. (i) and (ii), we have
\(V_{A}=V_{B}>V_{C}\)
\(V_{A}=V_{B}...(i)\)
Electric field and electric potential are related as
\(E=-\frac{d V}{d x} \Rightarrow V=-\int E d x\)
Which indicates that electric potential decreases in the direction of electric field, i.e.
\(V_{B}>V_{C}...(ii)\)
From Eqs. (i) and (ii), we have
\(V_{A}=V_{B}>V_{C}\)
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