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(D) When a dielectric slab of thickness $t$ is introduced between two point charges separated by distance $r$,the effective force $F$ is given by: $F

Vedclass · Accepted Answer

$100/81$

Vedclass · Answer

(D) When a dielectric slab of thickness $t$ is introduced between two point charges separated by distance $r$,the effective force $F$ is given by: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - t + t \sqrt{K})^2}$ Case $1$: $t = r/2$ and $K = 4$. $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - r/2 + (r/2) \sqrt{4})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r/2 + r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(3r/2)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4 q_1 q_2}{9 r^2} = \frac{q_1 q_2}{9 \pi \varepsilon_0 r^2}$. Case $2$: $t = r/3$ and $K = 9$. $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(r - r/3 + (r/3) \sqrt{9})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(2r/3 + r)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(5r/3)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{9 q_1 q_2}{25 r^2}$. Ratio $F_1/F_2$: $\frac{F_1}{F_2} = \left( \frac{q_1 q_2}{9 \pi \varepsilon_0 r^2} ight) / \left( \frac{9 q_1 q_2}{100 \pi \varepsilon_0 r^2} ight) = \frac{1}{9} 	imes \frac{100}{9} = \frac{100}{81}$.

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