AP EAMCET · PHYSICS · Capacitance
A capacitor of capacitance \(2 \mu \mathrm{~F}\) is charged to 50 V and then disconected from the source. Later the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by \(25 \%\) of its initial value, then the dielectric constant of the dieletric material is
- A \(\frac{2}{3}\)
- B \(\frac{4}{3}\)
- C \(\frac{3}{4}\)
- D \(\frac{3}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(U_f = U_i - 0.25 U_i = 0.75 U_i\) \(U_f = \frac{U_i}{k}\) \(\frac{U_i}{k} = 0.75 U_i\) \(k = \frac{1}{0.75} = \frac{4}{3}\)
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