A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a di

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2. Electric Potential and Capacitance

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A parallel plate capacitor is made of two circular plates separated by a distance $5\ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4$ $ Vm^{-1}$ the charge density of the positive plate will be close to

A

$3 \times 10^{-7} $ $Cm^{-2}$

B

$3 \times 10^4$ $ Cm^{-2}$

C

$6 \times 10^4 $ $Cm^{-2}$

D

$6 \times 10^{-7}$ $Cm^{-2}$

(JEE MAIN-2014)

Solution

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

$E=\frac{\sigma}{K \varepsilon_{0}}$

Then, charge density

${\sigma=K \varepsilon_{0} E}$

${=2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4} \approx 6 \times 10^{-7} \,\mathrm{C} / \mathrm{m}^{2}}$

Standard 12

Physics

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