MHT CET · Physics · Electrostatics
The electric potential at the centre of two concentric half rings of radii \(R_1\) and \(R_2\), having same linear charge density ' \(\lambda\) ' is ( \(\varepsilon_0=\) permittivity of free space \()\)
- A \(\frac{2 \lambda}{\varepsilon_0}\)
- B \(\frac{\lambda}{2 \varepsilon_0}\)
- C \(\frac{\lambda}{4 \varepsilon_0}\)
- D \(\frac{\lambda}{\varepsilon_0}\)
Answer & Solution
Correct Answer
(B) \(\frac{\lambda}{2 \varepsilon_0}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{V}_1=\frac{1}{4 \pi \varepsilon_0} \times \frac{\lambda\left(\pi \mathrm{R}_1\right)}{\mathrm{R}_1} \\ & \mathrm{~V}_2=\frac{1}{4 \pi \varepsilon_0} \times \frac{\lambda\left(\pi \mathrm{R}_2\right)}{\mathrm{R}_2} \\ & \mathrm{~V}_{\mathrm{net}}=\mathrm{V}_1+\mathrm{V}_2=\frac{1}{4 \pi \varepsilon_0} \times \pi \lambda \times 2=\frac{\lambda}{2 \varepsilon_0}\end{aligned}\)
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