CUET Preparation Today
Consider a capacitor circuit charged by a battery to a potential difference $V_O$ as shown in the figure.
When a dielectric slab is inserted into both the capacitors, new potential difference will be |
$V_0$ $\frac{V_0}{K}$ $\frac{8V_0}{5K}$ $\frac{8V_0}{K}$ |
$\frac{8V_0}{5K}$ |
The correct answer is option (3) : $\frac{8V_0}{5K}$
On inserting dielectric charge on the combination of capacitor remains same $q=cV$ $4C_0V_0=\left(KC_0+\frac{3C_0K}{2}\right)V_1$ $V_1=\frac{8V_0}{5K}$ |
