A force F acts between sodium and chlorine ions of salt (sodium chloride) when put 1,cm apart in air

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2. Electric Potential and Capacitance

easy

A force $F$ acts between sodium and chlorine ions of salt (sodium chloride) when put $1\,cm$ apart in air. The permittivity of air and dielectric constant of water are ${\varepsilon _0}$ and $K$ respectively. When a piece of salt is put in water electrical force acting between sodium and chlorine ions $1\,cm$ apart is

$\frac{F}{K}$

$\frac{{FK}}{{{\varepsilon _0}}}$

$\frac{F}{{K{\varepsilon _0}}}$

$\frac{{F{\varepsilon _0}}}{K}$

Solution

(a) When put $1$ $cm$ apart in air, the force between $Na$ and $Cl$ ions = $F$. When put in water, the force between $Na$ and $Cl$ ions $ = \frac{F}{K}$

Standard 12

Physics

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(AIIMS-2001)

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medium

(AIIMS-2008)

Solution

Similar Questions

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is $A\;metr{e^2}$ and the separation is $t$ $metre$. The dielectric constants are ${k_1}$ and ${k_2}$ respectively. Its capacitance in farad will be

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Assertion : A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant $K$ is introduced between the plates. The energy which is stored becomes $K$ times.

Reason : The surface density of charge onthe plate remains constant or unchanged.