AP EAMCET · PHYSICS · Electrostatics
The force between two point charges kept with a separation of 9 cm in air is 98 N. If a dielectric slab of constant 4, thickness 6 cm and another dielectric slab of constant 9, thickness 3 cm are introduced between the two charges, then the new force becomes
- A 18 N
- B 36 N
- C 49 N
- D 84 N
Answer & Solution
Correct Answer
(A) 18 N
Step-by-step Solution
Detailed explanation
\(r_0 = 9 \, \text{cm}\) \(F_0 = 98 \, \text{N}\) \(r_{eff} = t_1 \sqrt{K_1} + t_2 \sqrt{K_2}\) \(r_{eff} = 6 \sqrt{4} + 3 \sqrt{9}\) \(r_{eff} = 6(2) + 3(3)\) \(r_{eff} = 12 + 9 = 21 \, \text{cm}\) \(F = F_0 \left(\frac{r_0}{r_{eff}}\right)^2\)…
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