KCET · Physics · Electrostatics
A uniform electric field in the plane of the paper as shown. Here \(A, B, C\) and \(D\) are the ponits on the circle. \(V_{1}, V_{2}, V_{3}\) and \(V_{4}\) are the potentials at those points respectively. Then
- A \(V_{A}=V_{C}, V_{B}=V_{D}\)
- B \(V_{A}=V_{C}, V_{B}>V_{D}\)
- C \(V_{A}>V_{C}, V_{B}>V_{D}\)
- D \(V_{A}=V_{B}, V_{C}=V_{D}\)
Answer & Solution
Correct Answer
(D) \(V_{A}=V_{B}, V_{C}=V_{D}\)
Step-by-step Solution
Detailed explanation
The field is uniform in the plane of the paper. We know that in uniform electric field, the value of potentials is same for those points which are at equal distances. Therefore, \(V_{A}=V_{B}\) (Both points \(A\) and \(B\) lies on opposite sides on the circumference) and \(V_{C}>V_{D}\) (As point \(C\) lies earlier than point \(D\) on the circumference).
\(V_{A}=V_{B}, \quad V_{C}>V_{D}\)
\(V_{A}=V_{B}, \quad V_{C}>V_{D}\)
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