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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor with area \(200\,cm^2\) and separation between the plates \(1.5\,cm\), is connected across a battery of \(emf\) \(V\). If the force of attraction between the plates is \(25\times10^{-6}\,N\), the value of \(V\) is approximately........\(V\) \(\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}\,\frac{{{C^2}}}{{N{m^2}}}} \right)\)
- A \(150\)
- B \(100\)
- C \(250\)
- D \(300\)
Answer & Solution
Correct Answer
(C) \(250\)
Step-by-step Solution
Detailed explanation
Given area of Parallel plate capacitor, \(A=\) \(200\, \mathrm{cm}^{2}\) Separation between the plates, \(d=1.5\, \mathrm{cm}\) Force of attraction between the plates, \(F=\) \(25 \times 10^{-6}\, \mathrm{N}\) \(F=Q E\) \(F=\frac{Q^{2}}{2 A \epsilon_{0}} \quad\) (\(E\) due to…
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