JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \(6 \times 10^{-7}\) \(Cm^{-2}\)
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 \({\sigma=K \varepsilon_{0} E}\)…
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