Initial charge stored $=50\mu\text{c}$
Let the dielectric constant of the material induced be ‘k’.
Now, when the extra charge flown through battery is 100.
So, net charge stored in capacitor $=150\mu\text{c}$
Now, $\text{C}_1=\frac{\in_0\text{A}}{\text{d}}$ or $\frac{\text{q}_1}{\text{V}}=\frac{\in_0\text{A}}{\text{d}}\ \dots(1)$
$\text{C}_2=\frac{\in_0\text{Ak}}{\text{d}}$ or $\frac{\text{q}_2}{\text{V}}=\frac{\in_0\text{Ak}}{\text{d}}\ \dots(2)$
Dividing (1) and (2) we get $\frac{\text{q}_1}{\text{q}_2}=\frac{1}{\text{k}}$
$\Rightarrow\frac{50}{150}=\frac{1}{\text{k}}$
$\Rightarrow\text{k}=3$
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