JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor filled with a medium of dielectric constant \(10\) , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant \(15\) . Then the energy of capacitor will ......................
- A increase by \(50 \%\)
- B decrease by \(15 \%\)
- C increase by \(25 \%\)
- D increase by \(33 \%\)
Answer & Solution
Correct Answer
(A) increase by \(50 \%\)
Step-by-step Solution
Detailed explanation
\(E \Rightarrow \frac{1}{2}( KC ) v ^{2}\) \(\therefore \%\) change \(\Rightarrow \frac{\frac{1}{2} K _{2} CV ^{2}-\frac{1}{2} K _{1} CV ^{2}}{\frac{1}{2} K _{1} CV ^{2}}=\frac{ K _{2}- K _{1}}{ K _{1}} \times 100\) \(\Rightarrow \frac{15-10}{10} \times 100=50 \%\)
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