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

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Answer & Solution

Correct Answer

(D) \(6 \times 10^{-7}\) \(Cm^{-2}\)

Step-by-step Solution

Detailed explanation

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

\({\sigma=K \varepsilon_{0} E}\)

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

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