Question 612: Electrostatics & Capacitance

Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $	heta$ 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 $=1 \mathrm{~g} / \mathrm{cc}$ )

Accepted Answer

In air $	an \frac{	heta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 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
$$
\begin{aligned}
& \varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}ight] \\
& \varepsilon_{\mathrm{r}}=3
\end{aligned}
$$

Electrostatics & Capacitance

Integer / Numeric Answer Type