JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor has plates of area \(A\) separated by distance \(d\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(\mathrm{k}(\mathrm{x})=\mathrm{K}(1+\alpha \mathrm{x})\) where \(\mathrm{x}\) is the distance measured from one of the plates. If \((\alpha \text {d)}<<1,\) the total capacitance of the system is best given by the expression
- A \(\frac{\mathrm{AK} \varepsilon_{0}}{\mathrm{d}}\left(1+\frac{\alpha \mathrm{d}}{2}\right)\)
- B \(\frac{\mathrm{A} \varepsilon_{0} \mathrm{K}}{\mathrm{d}}\left(1+\left(\frac{\alpha \mathrm{d}}{2}\right)^{2}\right)\)
- C \(\frac{\mathrm{A} \varepsilon_{0} \mathrm{K}}{\mathrm{d}}\left(1+\frac{\alpha^{2} \mathrm{d}^{2}}{2}\right) \)
- D \( \frac{\mathrm{AK} \varepsilon_{0}}{\mathrm{d}}(1+\alpha \mathrm{d})\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{AK} \varepsilon_{0}}{\mathrm{d}}\left(1+\frac{\alpha \mathrm{d}}{2}\right)\)
Step-by-step Solution
Detailed explanation
As \(\mathrm{K}\) is variable we take a plate element of Area \(A\) and thickness \(dx\) at distance \(\mathrm{x}\) Capacitance of element \(\mathrm{d} \mathrm{C}=\frac{(\mathrm{A}) \mathrm{K}(1+\alpha \mathrm{x}) \varepsilon_{0}}{\mathrm{dx}}\) Now all such elements are is…
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