Question

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 $=10$ )

Hint:

In parallel combinations, the total charge is the sum of individual charges, and the total capacitance is the sum of individual capacitances.

Answer 1

Correct Answer: 55

Given : Charge on C₁ = KCE
Charge on C₂ = CE
When they are connected in parallel charge will be equally divided so charge on one capacitor is
\(q=\frac{K+1}{2}CV\)
\(\therefore\) \(V=\frac{q}{KC}\)
\(=\frac{K+1}{2K}=55 V\)
so, the correct answer is 55.

Answer 2

Step 1: The charge on \( C_1 \) is given by:
\( Q_1 = K C E \)
where \( K \) is the dielectric constant and \( E \) is the electric field.

Step 2: The charge on \( C_2 \) is given by:
\( Q_2 = C E \)

Step 3: When they are connected in parallel, the charge is equally divided, so the charge on one capacitor is:
\( q = \frac{K + 1}{2} C V \)

Step 4: Now, the voltage across the capacitors is given by:
\( V = \frac{q}{C K} = \frac{K + 1}{2K} = 55 \, \text{V} \)

Correct Answer: 55 V