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(B) The electrostatic force between two charges $q$ in a medium with dielectric constant $K$ is given by $F = \frac{1}{4 \pi \epsilon_0 K} \frac{q^2}{

Vedclass · Accepted Answer

$3.2$

Vedclass · Answer

(B) The electrostatic force between two charges $q$ in a medium with dielectric constant $K$ is given by $F = \frac{1}{4 \pi \epsilon_0 K} \frac{q^2}{r^2}$.In ice at $-10^{\circ} C$,the force is $F_i = \frac{1}{4 \pi \epsilon_0 K_i} \frac{q^2}{(25 	imes 10^{-2})^2}$.In water at $0^{\circ} C$,the force is $F_w = \frac{1}{4 \pi \epsilon_0 (80)} \frac{q^2}{(5 	imes 10^{-2})^2}$.Since the interaction force remains the same $(F_i = F_w)$,we equate the two expressions:$\frac{1}{K_i (25)^2} = \frac{1}{80 (5)^2}$.$K_i = 80 	imes \frac{5^2}{25^2} = 80 	imes \frac{25}{625} = 80 	imes \frac{1}{25} = 3.2$.Therefore,the dielectric constant of ice is $3.2$.

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