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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
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The correct answer is:
$V_1>V_2, C_1>C_2$
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
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
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