JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: \(K(x) = K_0 + \lambda x\) ( \(\lambda =\) constant) The capacitance \(C,\) of the capacitor, would be related to its vacuum capacitance \(C_0\) for the relation
- A \(C\, = \,\frac{{\lambda d}}{{\ln \,(1 + {K_0}\lambda d)}}{C_0}\)
- B \(C\, = \,\frac{{\lambda }}{{d.\ln \,(1 + {K_0}\lambda d)}}{C_0}\)
- C \(C\, = \,\frac{{\lambda d}}{{\ln \,(1 + \lambda d/{K_0})}}{C_0}\)
- D \(C\, = \,\frac{\lambda }{{d.\ln \,(1 + {K_0}/\lambda d)}}{C_0}\)
Answer & Solution
Correct Answer
(C) \(C\, = \,\frac{{\lambda d}}{{\ln \,(1 + \lambda d/{K_0})}}{C_0}\)
Step-by-step Solution
Detailed explanation
The value of diclectric constant is given as \(\mathrm{K}=\mathrm{K}_{0}+\lambda \mathrm{x}\) And, \(\mathrm{V}=\int_{0}^{\mathrm{d}} \mathrm{Edr}\) \(\mathrm{v}=\int_{0}^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}\)…
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