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Step-by-Step Solution
Step 1: Understand the Problem
We have a combination of parallel plate capacitors maintained at a certain potential difference. Each capacitor initially has a certain plate separation and no dielectric. When a dielectric slab of thickness 3 mm is introduced between the plates of all the capacitors, the plate separation is increased by 2.4 mm to maintain the same potential difference. We need to find the dielectric constant $K$ of the slab.
The diagram (given below) shows schematically how the slab is placed:
Step 2: Recall the Formula for Capacitance of a Parallel Plate Capacitor
For a parallel plate capacitor of plate area $A$ and plate separation $d$ (with vacuum or air between plates), the capacitance is
$$
C = \frac{\varepsilon_0 \, A}{d}.
$$
If a dielectric of dielectric constant $K$ and thickness $t$ is introduced between the plates uniformly, the effective capacitance changes due to the change in effective separation.
Step 3: Identify the Original and Modified Configurations
• Original situation (without dielectric): Each capacitor has capacitance
$$
C_1 = \frac{\varepsilon_0 A}{3}.
$$
(Here, 3 mm is understood to be the separation for the original capacitor or is related to the original geometry.)
• After introducing the dielectric: The slab of dielectric constant $K$ and thickness 3 mm is inserted, and the plate separation is increased by 2.4 mm. This effectively creates two regions in each capacitor: one filled with the dielectric of thickness 3 mm and the other with air separation of 2.4 mm. The combined effect in total thickness is $(3 + 2.4) = 5.4\text{ mm}$, but realized as two layers in series:
$$
C_\text{new for one capacitor} = \left(\frac{K \, \varepsilon_0 \, A}{3}\right) \quad \text{in series with} \quad \left(\frac{\varepsilon_0 \, A}{2.4}\right).
$$
Step 4: Series Combination Condition for Capacitance
Because, after modification, each capacitor effectively behaves like two capacitors in series (one part with dielectric, one part with air), the overall capacitance for the single modified capacitor $C_1^\prime$ is given by the series formula:
$$
\frac{1}{C_1^\prime}
= \frac{1}{\frac{K \, \varepsilon_0 \, A}{3}}
+ \frac{1}{\frac{\varepsilon_0 \, A}{2.4}}.
$$
Meanwhile, to maintain the same potential difference as initially, the capacitance for each original capacitor $C_1 = \frac{\varepsilon_0 A}{3}$ should equal the new effective capacitance $C_1^\prime$.
Step 5: Equate the Original and New Capacitances
We set
$$
\frac{\varepsilon_0 A}{3}
= C_1^\prime
= \left[\frac{1}{\frac{K \, \varepsilon_0 \, A}{3}}
+ \frac{1}{\frac{\varepsilon_0 \, A}{2.4}}\right]^{-1}.
$$
Simplifying inside the brackets,
$$
\frac{1}{\frac{K \, \varepsilon_0 \, A}{3}} = \frac{3}{K \, \varepsilon_0 \, A},
\quad
\frac{1}{\frac{\varepsilon_0 \, A}{2.4}} = \frac{2.4}{\varepsilon_0 \, A}.
$$
Hence,
$$
\frac{1}{C_1^\prime}
= \frac{3}{K \, \varepsilon_0 \, A} + \frac{2.4}{\varepsilon_0 \, A}
= \frac{3 + 2.4K}{K \, \varepsilon_0 \, A}.
$$
So
$$
C_1^\prime = \frac{K \, \varepsilon_0 \, A}{3 + 2.4K}.
$$
Setting $C_1^\prime = \frac{\varepsilon_0 A}{3}$, we get
$$
\frac{\varepsilon_0 A}{3} = \frac{K \, \varepsilon_0 \, A}{3 + 2.4K}.
$$
Step 6: Solve for the Dielectric Constant $K$
Canceling $\varepsilon_0 A$ from both sides, we get
$$
\frac{1}{3} = \frac{K}{3 + 2.4K}.
$$
Cross-multiply:
$$
3 + 2.4K = 3K.
$$
Rearrange to solve for $K$:
$$
3 = 3K - 2.4K = 0.6K
\quad \Longrightarrow \quad
K = \frac{3}{0.6} = 5.
$$
Step 7: Final Answer
The dielectric constant of the slab is 5.
