Capacitor With Dielectrics: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Capacitor With Dielectrics. Understanding the underlying formula is key to arriving at the correct answer below:

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:

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