KCET · Physics · Capacitance
A parallel plate capacitor of capacitance \(C_1\) with a dielectric slab in between its plates is connected to a battery. It has a potential difference \(V_1\) across its plates. When the dielectric slab is removed, keeping the capacitor connected to the battery, the new capacitance and potential difference are \(C_2\) and \(V_2\) respectively, Then
- A \(V_1=V_2, C_1<C_2\)
- B \(V_1>V_2, C_1>C_2\)
- C \(V_1<V_2, C_1>C_2\)
- D \(V_1=V_2, C_1>C_2\)
Answer & Solution
Correct Answer
(B) \(V_1>V_2, C_1>C_2\)
Step-by-step Solution
Detailed explanation
Capacitance when dielectric slab is
introduced.
\(C_1=\frac{k \varepsilon_0 A}{d}\)
As we know, \(C=\frac{Q}{V_1}\)
But when dielectric is removed, \(C_1=\frac{\varepsilon_0 A}{d}=\frac{Q}{V_2}\)
Clearly, \(C_2<C_1\)
and \(\quad \frac{Q}{V_2}<\frac{Q}{V_1} \Rightarrow V_1>V_2\)
introduced.
\(C_1=\frac{k \varepsilon_0 A}{d}\)
As we know, \(C=\frac{Q}{V_1}\)
But when dielectric is removed, \(C_1=\frac{\varepsilon_0 A}{d}=\frac{Q}{V_2}\)
Clearly, \(C_2<C_1\)
and \(\quad \frac{Q}{V_2}<\frac{Q}{V_1} \Rightarrow V_1>V_2\)
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