Radii of inner and outer spheres of a condenser are 9,cm and 10,cm respectively. If dielectric const

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2. Electric Potential and Capacitance

medium

The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$

$180$

$30$

$18$

$90$

Solution

(b) Given system is a spherical capacitor
So capacitance of system $C = K \times 4\pi {\varepsilon _0}\left[ {\frac{{{r_1}{r_2}}}{{{r_2} – {r_1}}}} \right]$
$ = \frac{6}{{9 \times {{10}^9}}}\left[ {\frac{{9 \times 10}}{1}} \right] \times {10^{ – 2}} = 6 \times {10^{ – 10}}\,Farad$
Now potential of inner sphere will be equal to potential difference of the capacitor. So $V = \frac{q}{C} = \frac{{18 \times {{10}^{ – 9}}}}{{6 \times {{10}^{ – 10}}}} = 30\,V$

Standard 12

Physics

The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out

normal

A parallel plate capacitor with air as medium between the plates has a capacitance of $10\,\mu F$. The area of capacitor is divided into two equal halves and filled with two media as shown in the figure having dielectric constant ${k_1} = 2$and ${k_2} = 4$. The capacitance of the system will now be…….$\mu F$

medium

Assertion : The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates.

Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.

easy

(AIIMS-2009)

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medium

A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :

hard

(JEE MAIN-2022)

Solution

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The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out

Assertion : The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates. Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.

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Assertion : The electrostatic force between the plates of a charged isolated capacitor decreases when dielectric fills whole space between plates.

Reason : The electric field between the plates of a charged isolated capacitance increases when dielectric fills whole space between plates.