Two capacitors of capacitance $10\mu\text{F}$ and $20 \mu\text{F}$ are connected in series with a 6 V battery. After the capacitors are fully charged, a slab of dielectric constant (K) is inserted between the plates of the two capacitors. How will the following be affected after the slab is introduced:
- The electric field energy stored in the capacitors.
- The charges on the two capacitors.
- The potential difference between the plates of the capacitors.
CBSE OUTSIDE DELHI - SET 2 BHUBNESHWAR 2015
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The capacitance of both the capacitors increases by a factor K.
Alternate Answer
- New Electric field energy values are: = $\frac{1}{2}\text{k}(\text{c}_{1}\text{v}^{2}_{1}) \text{ and}\frac{1}{2}\text{k}(\text{c}_{2}\text{v}^{2}_{1}).$
- New charges are: =$\frac{1}{2}\text{KC}_{1}\text{V}_{1}\text{ and }\frac{1}{2}\text{KC}_{2}\text{V}_{2}.$
- New P.D values are: $V_1$ and $V_2$
Alternate Answer
- New Electric field energy values are: $\frac{1}{2}\frac{\text{Q}^{2}}{\text{KC}_{1}} \text{and } \frac{1}{2}\frac{\text{Q}^{2}}{\text{KC}_{2}}.$
- New charges are: Q and Q as before.
- New P.D values are: $\frac{\text{Q}}{\text{KC}_{1}} \text{and }\frac{\text{Q}}{\text{KC}_{2}}.$
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