Q160 / 180 · KCET 2023 · Physics
A par allel 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$V_1=V_2, C_1<C_2$
- Correct answerB$V_1>V_2, C_1>C_2$
- C$V_1<V_2, C_1>C_2$
- D$V_1=V_2, C_1>C_2$
Solution
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 $\quad \frac{Q}{V_2}<\frac{Q}{V_1} \Rightarrow V_1>V_2$
Hence, the correct answer is option: 'B'.