Two identical charged spheres are suspended b | vedclass

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Question

In air $	an \frac{	heta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 \mathrm{r}^2 m g}$

In water $	an \frac{	heta}{2}=\frac{\mathrm{F}^{\pr

Vedclass · Accepted Answer

$3$

Vedclass · Answer

In air $	an \frac{	heta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 \mathrm{r}^2 m g}$

In water $	an \frac{	heta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{	ext {eff }}}$

Equate both equations

$\varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}ight]$

$\varepsilon_{\mathrm{T}}=3$

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