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

To solve this problem, we need to analyze the effects of inserting dielectric slabs into the capacitors and then determine the common potential when they are connected in parallel. Let's break down the solution step by step:

First, consider capacitor C₁, which is kept connected to the 100 V DC source during the insertion of the dielectric slab. The capacitance of the capacitor with the dielectric, C_1f, is calculated as:

C_1f=K×C=10×10μF=100μF

Since C₁ is connected to the source, its voltage remains constant at 100 V. Thus, the charge Q₁ on C₁ becomes:

Q₁=C_1f×V=100μF×100V=10000μC

Now consider capacitor C₂, which is disconnected during the insertion of the dielectric slab. The initial charge on C₂ when it is disconnected is:

Q_2i=C×V=10μF×100V=1000μC

After inserting the dielectric, the new capacitance C_2f becomes:

C_2f=K×C=10×10μF=100μF

The charge on C₂ remains the same as it was before the dielectric was inserted since it is isolated. Therefore, the new voltage across C₂, V₂, is given by:

V₂=Q_2i/C_2f=1000μC/100μF=10V

When the two capacitors are connected in parallel, charges distribute to balance the potential difference across both. The total charge Q_t is:

Q_t=Q₁+Q_2i=10000μC+1000μC=11000μC

The equivalent capacitance C_eq is:

C_eq=C_1f+C_2f=100μF+100μF=200μF

The common potential V_common is calculated as:

V_common=Q_t/C_eq=11000μC/200μF=55V

This common potential falls within the specified range of 55 V. Thus, the final answer is 55 V.