The plates of a parallel plate capacitor are separated by a distance '\(d\)' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by \(50%\). The thickness of the dielectric slab is
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Partial dielectric behaves like series combination of two capacitors.
Step 1: Understanding the Concept:
Introducing a dielectric slab between the plates of a capacitor increases its capacitance.
The new capacitance depends on the dielectric constant of the slab and its thickness relative to the plate separation. Step 2: Key Formulas or Approach:
Initial capacitance with air: \( C_0 = \frac{\varepsilon_0 A}{d} \).
Capacitance with a dielectric slab of thickness \( t \) and dielectric constant \( K \): \( C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \). Step 3: Detailed Explanation:
The problem states that the capacity increases by \( 50% \).
This means the new capacitance is \( C = C_0 + 0.5 C_0 = 1.5 C_0 = \frac{3}{2} C_0 \).
Substitute the expression for \( C_0 \):
\[ C = \frac{3}{2} \left( \frac{\varepsilon_0 A}{d} \right) \]
We are given the dielectric constant \( K = 3 \).
Now, use the formula for capacitance with a partially filled dielectric:
\[ \frac{\varepsilon_0 A}{d - t + \frac{t}{3}} = \frac{3}{2} \frac{\varepsilon_0 A}{d} \]
Cancel the common term \( \varepsilon_0 A \) from both sides:
\[ \frac{1}{d - t + \frac{t}{3}} = \frac{3}{2d} \]
Simplify the denominator on the left side: \( -t + \frac{t}{3} = \frac{-3t + t}{3} = \frac{-2t}{3} \).
\[ \frac{1}{d - \frac{2t}{3}} = \frac{3}{2d} \]
Cross-multiply to solve for \( t \):
\[ 2d = 3\left(d - \frac{2t}{3}\right) \]
Expand the right side:
\[ 2d = 3d - 3\left(\frac{2t}{3}\right) \]
\[ 2d = 3d - 2t \]
Rearrange to isolate \( t \):
\[ 2t = 3d - 2d \]
\[ 2t = d \]
\[ t = \frac{d}{2} \]
Step 4: Final Answer:
The thickness of the dielectric slab is \( \frac{d}{2} \).