Capacitance Question 5
Q5 - 31 January - Shift 2
Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a 100 V D.C. source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be V. (Assuming Dielectric constant $=10$ )
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Answer: (55)
Solution:
Formula: Capacitor with Dielectric
$\frac{1(90 \mathrm{mc})^2}{25(5 \mathrm{~s})}-2 \times \frac{1}{2} \frac{(45 \mathrm{mc})^2}{900 \mathrm{uF}}\left|\mathrm{U}=\frac{\mathrm{Q}^2}{2 \mathrm{c}}\right|$
$=2.25$ Joule
