TS EAMCET · Physics · Electrostatics
A thin non-conducting ring of radius \(r\) has a linear charge density \(\lambda=\lambda_0 \cos \phi\), where \(\lambda_0\) is a constant and \(\phi\) is the azimuthal angle. The magnitude of the electric field strength at the centre of the ring is
- A \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda_0}{r}\)
- B \(\frac{1}{2 \pi \varepsilon_0} \frac{\lambda_0}{r}\)
- C \(\frac{\lambda_0}{4 \varepsilon_0 r}\)
- D \(\frac{\lambda_0}{2 \varepsilon_0 r}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda_0}{4 \varepsilon_0 r}\)
Step-by-step Solution
Detailed explanation
Given that, linear charge density of non-conducting ring \(\lambda=\lambda_0 \cos \phi\), and Azimuthal angle \(=\phi\) Consider two elements of length \((d l=R d \phi)\) are taken symmetrically at angle \(\phi\) on both sides as shown in figure. Here, \(d E_1=d E_2\) and…
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