NEET Physics Question - Capacitors - Capacitor With Dielectrics

Two identical parallel plate capacitors A and B are connected in series with a battery of 100 V. A slab of dielectric constant K = 3 is inserted between the plates of capacitor A. Then, the potential difference across the capacitors will be, respectively:

25 V, 75 V

75 V, 25 V

20 V, 80 V

50 V, 50 V

Solution:

Here’s a shorter solution:

Given:

$V = 100 \, \text{V}$ (total battery voltage)
Dielectric constant $K = 3$ for capacitor $A$
Identical capacitors $A$ and $B$

Step 1: Capacitance

$C_A = 3C_0$ (because of the dielectric in $A$ )
$C_B = C_0$ (no dielectric in $B$ )

Step 2: Total capacitance in series:

$\frac{1}{C_{\text{eq}}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{3C_0} + \frac{1}{C_0} = \frac{4}{3C_0}$ $C_{\text{eq}} = \frac{3C_0}{4}$

Step 3: Voltage division:

The voltage is divided in proportion to the inverse of capacitances:

$V_A = \frac{C_B}{C_A + C_B} \times 100 = \frac{C_0}{3C_0 + C_0} \times 100 = \frac{1}{4} \times 100 = 25 \, \text{V}$ $V_B = 100 - 25 = 75 \, \text{V}$

Final Answer:

$V_A = 25 \, \text{V}$
$V_B = 75 \, \text{V}$

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