KCET · Physics · Electrostatics
Electric field at a distance \(r\) from an infinitely long uniformly charged straight conductor, having linear charge density \(\lambda\) is \(E_1\). Another uniformly charged conductor having same linear charge density \(\lambda\) is bent. into a semicircle of radius \(r\). The electric field at its centre is \(E_2\). Then
- A \(E_2=\pi r E_1\)
- B \(E_2=\frac{E_1}{r}\)
- C \(E_1=E_2\)
- D \(E_1=\pi r E_2\)
Answer & Solution
Correct Answer
(C) \(E_1=E_2\)
Step-by-step Solution
Detailed explanation
Electric field due to infinitely charged long
wire, \(E_1=\frac{2 k \lambda}{r}\).
Electric field at the centre of semicircular ring,
\(E_2=\frac{2 k \lambda}{r}\)
Therefore, \(E_1=E_2\)
wire, \(E_1=\frac{2 k \lambda}{r}\).
Electric field at the centre of semicircular ring,
\(E_2=\frac{2 k \lambda}{r}\)
Therefore, \(E_1=E_2\)
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