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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2024
  • Hard
  • Numerical

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle \(\theta\) with each other. When suspended in water the angle remains the same. If density of the material of the sphere is \(1.5 \mathrm{~g} / \mathrm{cc}\), the dielectric constant of water will be _______. (Take density of water \(=1 \mathrm{~g} / \mathrm{cc}\) )

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Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

In air \(\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 \mathrm{r}^2 m g}\)

In water \(\tan \frac{\theta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{\text {eff }}}\)

Equate both equations

\(\varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}\right]\)

\(\varepsilon_{\mathrm{T}}=3\)

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