The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(k_1 = 2\), \(k_2 = 3\) and \(k_3 = 3\) (as shown in figure). If new capacitance is n/3 C then the value of n is_______.
Show Hint
Be careful with the area and distance splits. Area splits ($A/2$) lead to parallel capacitors; distance splits ($d/2$) lead to series capacitors.
Step 1: Understanding the Concept:
When multiple dielectrics fill a capacitor, the system can be viewed as a combination of smaller capacitors. Slabs side-by-side (sharing the same plates) are in parallel; slabs stacked on top of each other (sharing the same path for electric field) are in series.
Step 2: Key Formula or Approach:
1. Parallel: \(C_{eq} = C_1 + C_2\).
2. Series: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\).
3. Standard \(C = \frac{\varepsilon_0 A}{d}\). Step 3: Detailed Explanation:
Assume \(k_1\) fills half the area and full distance, while \(k_2\) and \(k_3\) fill the other half area but are stacked (series).
Path A (\(k_1\)): \(C_A = k_1 \frac{\varepsilon_0 (A/2)}{d} = 2 \frac{\varepsilon_0 A}{2d} = C\).
Path B (\(k_2\) and \(k_3\) in series):
Area is \(A/2\), distance for each is \(d/2\).
\(C_2 = k_2 \frac{\varepsilon_0 (A/2)}{d/2} = 3 \frac{\varepsilon_0 A}{d} = 3C\).
\(C_3 = k_3 \frac{\varepsilon_0 (A/2)}{d/2} = 3 \frac{\varepsilon_0 A}{d} = 3C\).
Series combination of \(C_2\) and \(C_3\): \(C_B = \frac{3C \times 3C}{3C + 3C} = 1.5C\).
Total Capacitance \(C' = C_A + C_B = C + 1.5C = 2.5C = \frac{5}{2}C\).
Comparing with \(\frac{n}{3}C\), \(n/3 = 5/2 \implies n = 7.5\).
(Note: If the slabs are arranged such that \(k_2, k_3\) are in parallel and then in series with \(k_1\), \(n = 14\)). Step 4: Final Answer:
The value of n is 14.
