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Two point charges are placed at separati...

Two point charges are placed at separation `d` in vacuum and the force between them is `F`. Now a dielectric slab of thickness `t = d//3` and dielectric constant `K` is placed between the charges and the force becomes `9F//25`. Find the value of `K`.

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To solve the problem step by step, we will use the concepts of electrostatics, particularly Coulomb's law and the effect of a dielectric on the force between two charges. ### Step 1: Understanding the Initial Condition Initially, we have two point charges \( q_1 \) and \( q_2 \) separated by a distance \( d \) in a vacuum. The force between them is given by Coulomb's law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{d^2} \] ...
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Knowledge Check

  • Two points charges Q_(1) and Q_(2) placed at separation d in vacuum and force acting between them is F . Now a dielectric slab of thickness d//2 and dielectric constant K = 4 is placed between them. The new force between the charges will be

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    `(4 F)/(9)`
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  • Two similar point charges q_1 and q_2 are placed at a distance r apart in the air. The force between them is F_1 . A dielectric slab of thickness t(ltr) and dielectric constant K is placed between the charges . Then the force between the same charge . Then the fore between the same charges is F_2 . The ratio is

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