JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor with plate area \(A\) and plate separation \(d =2 \,m\) has a capacitance of \(4 \,\mu F\). The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant \(K =3\) (as shown in figure) will be .........\( \mu \,F\)

- A \(2\)
- B \(32\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(C _{\text {original }}=\frac{ A \varepsilon_{0}}{ d }\) \(C _{1}=\frac{ A \varepsilon_{0}}{ d / 2}=\frac{2 A \varepsilon_{0}}{ d }= C\) \(C _{2}=\frac{ KA \varepsilon_{0}}{ d / 2}=\frac{2 KA \varepsilon_{0}}{ d }=\frac{6 A \varepsilon_{0}}{ d }=3 C\) \(C_{1}\) and \(C_{2}\)…
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