Question Number 226907 by fantastic2 last updated on 18/Dec/25
$${two}\:{small}\:{balls}\:{are}\:{hung}\:{from}\:{a}\:{point} \\ $$$$\left({same}\:{mass},\:{same}\:{charge}\:{and}\:{rope}\:{length}\:{are}\:{same}\right) \\ $$$${the}\:{two}\:{strings}\:{make}\:{an}\:{angle}\:\mathrm{30}^{\mathrm{0}} \\ $$$${when}\:{immersed}\:{in}\:{a}\:{liquid}\:{of}\:\rho=\mathrm{0}.\mathrm{8}{g}/{cc} \\ $$$${the}\:{angle}\:{remains}\:{same}.\rho_{{ball}} =\mathrm{1}.\mathrm{6}{g}/{cc} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\:\kappa\left({dielectric}\:{const}.\right){of} \\ $$$${the}\:{liquid} \\ $$
Answered by mr W last updated on 23/Dec/25
Commented by mr W last updated on 23/Dec/25
$${in}\:{air}\:\left(\approx{vacuum}\right): \\ $$$${F}_{\mathrm{1}} ={mg}\:\mathrm{tan}\:\alpha_{\mathrm{1}} =\frac{{kq}^{\mathrm{2}} }{\left(\mathrm{2}{l}\:\mathrm{sin}\:\alpha_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$${in}\:{liquid}: \\ $$$${F}_{\mathrm{2}} =\left(\mathrm{1}−\frac{\rho_{{liquid}} }{\rho_{{ball}} }\right){mg}\:\mathrm{tan}\:\alpha_{\mathrm{2}} =\frac{{kq}^{\mathrm{2}} }{\kappa\left(\mathrm{2}{l}\:\mathrm{sin}\:\alpha_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{1}−\frac{\rho_{{liquid}} }{\rho_{{ball}} }\right)\frac{\mathrm{tan}\:\alpha_{\mathrm{2}} }{\mathrm{tan}\:\alpha_{\mathrm{1}} }=\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha_{\mathrm{1}} }{\kappa\:\mathrm{sin}^{\mathrm{2}} \:\alpha_{\mathrm{2}} } \\ $$$$\Rightarrow\kappa=\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha_{\mathrm{1}} \:\mathrm{tan}\:\alpha_{\mathrm{1}} }{\mathrm{sin}^{\mathrm{2}} \:\alpha_{\mathrm{2}} \:\mathrm{tan}\:\alpha_{\mathrm{2}} }×\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\rho_{{liquid}} }{\rho_{{ball}} }\right)} \\ $$$$\alpha_{\mathrm{1}} =\alpha_{\mathrm{2}} =\mathrm{30}°: \\ $$$$\kappa=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{0}.\mathrm{8}}{\mathrm{1}.\mathrm{6}}}=\mathrm{2} \\ $$
Commented by fantastic2 last updated on 23/Dec/25
$${correct}.\:{thank}\:{you} \\ $$