In figure shown, two parallel plate capacitors with fixed plates and connected to two batteries. Th…

Question

In figure shown, two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is same for the two capacitors. The plates are rectangular in shape with width b and lengths λ 1 and λ 2 , the separation between plates is d. The left half of the dielectric slab has a dielectric constant K 1 and the right half K 2 (K 2 > K 1 ). EMF of the right battery is greater then left battery. Neglecting any friction, find the extension in spring in equilibrium (spring is nonconducting) (ε 2 >ε 1 )

CGP EDU Academic Team · Accepted Answer

The correct answer is A. Step 1: Identify parameters:
- Capacitance of capacitor 1, C1 = \frac{K_1 \lambda_1 b}{d}
- Capacitance of capacitor 2, C2 = \frac{K_2 \lambda_2 b}{d}

Step 2: Find charge on each capacitor:
Using Q = CV,
- Charge on capacitor 1: Q1 = C1 * ε1 = \frac{K_1 \lambda_1 b \epsilon_1}{d}
- Charge on capacitor 2: Q2 = C2 * ε2 = \frac{K_2 \lambda_2 b \epsilon_2}{d}

Step 3: Determine force due to electric field:
Electric field in capacitor 1: E1 = \frac{Q1}{\epsilon_0 A} = \frac{K_1 \lambda_1 b \epsilon_1}{\epsilon_0 d}
Electric field in capacitor 2: E2 = \frac{Q2}{\epsilon_0 A} = \frac{K_2 \lambda_2 b \epsilon_2}{\epsilon_0 d}
Force on the spring: F = Q * E 
- F1 = Q1 * E1 = \frac{K_1 \lambda_1 b \epsilon_1}{d} * \frac{K_1 \lambda_1 b \epsilon_1}{\epsilon_0 d}
- F2 = Q2 * E2 = \frac{K_2 \lambda_2 b \epsilon_2}{d} * \frac{K_2 \lambda_2 b \epsilon_2}{\epsilon_0 d}

Step 4: Set up equilibrium condition:
In equilibrium, forces balance:
Kx = F2 - F1, where K is the spring constant and x is extension.
Solving for x will give us the extension in the spring. 
 Therefore, the extension in the spring in equilibrium can be derived from this setup.

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