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Question: Two similar point charges q<sub>1</sub> and q<sub>2</sub> are placed at a distance r, apart in air. ...

Two similar point charges q1 and q2 are placed at a distance r, apart in air. The force between them is F1. A dielectric slab of thickness t(< r) and dielectric constant K is placed between the charges. Then the force between the same charges is F2. The ratio F1/F2 is-

A

1

B

K

C

[r–t+tKr]2\left\lbrack \frac{r–t + t\sqrt{K}}{r} \right\rbrack^{2}

D

[rr–t+tK]2\left\lbrack \frac{r}{r–t + t\sqrt{K}} \right\rbrack^{2}

Answer

[r–t+tKr]2\left\lbrack \frac{r–t + t\sqrt{K}}{r} \right\rbrack^{2}

Explanation

Solution

F1= Kq1q2r2=q1q24πε0r2\frac { \mathrm { Kq } _ { 1 } \mathrm { q } _ { 2 } } { \mathrm { r } ^ { 2 } } = \frac { \mathrm { q } _ { 1 } \mathrm { q } _ { 2 } } { 4 \pi \varepsilon _ { 0 } \mathrm { r } ^ { 2 } } , F2 = q1q24πε0(r−t+tk)2\frac { q _ { 1 } q _ { 2 } } { 4 \pi \varepsilon _ { 0 } ( r - t + t \sqrt { k } ) ^ { 2 } }