KCET · Physics · Electrostatics
Electric field due to infinite, straight uniformly charged wire varies with distance \(r\) as
- A \(r\)
- B \(\frac{1}{r}\)
- C \(\frac{1}{r^{2}}\)
- D \(r^{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{r}\)
Step-by-step Solution
Detailed explanation
Electric field intensity due to infinity straight uniformly charged wire at some distance \(r\) is,
\(E=\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
where, \(\lambda=\) the surface charge density.
If \(\lambda, \pi\) and \(\varepsilon_{0}\) are constants, then
\(E \propto \frac{1}{r}\)
\(E=\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
where, \(\lambda=\) the surface charge density.
If \(\lambda, \pi\) and \(\varepsilon_{0}\) are constants, then
\(E \propto \frac{1}{r}\)
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