A parallel plate capacitor has charge $ 5 \times 10^{-6} \, C $. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $ 4 \times 10^{-6} \, C $ then the dielectric constant of the slab is ____.
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Correct Answer: 5
Approach Solution - 1
To solve the problem of finding the dielectric constant \( k \) of a slab in a parallel plate capacitor, we start by understanding the relationship between the free charge, induced charge, and the dielectric constant. The free charge on the capacitor, \( Q \), is given as \( 5 \times 10^{-6} \, \text{C} \). The induced charge on the dielectric, \( Q_i \), is \( 4 \times 10^{-6} \, \text{C} \).
Step 1: Understanding the Relationship
When a dielectric is inserted, it polarizes and reduces the effective electric field between the plates of a capacitor. The induced charge \( Q_i \) is related to the induced polarization caused by the dielectric.
The relationship between the total charge \( Q \), induced charge \( Q_i \), and the dielectric constant \( k \) is given by the equation:
\( Q = k \cdot (Q - Q_i) \)
Step 2: Solving for the Dielectric Constant, \( k \)
Rearrange the equation to express \( k \):
\( k = \frac{Q}{Q - Q_i} \)
Substitute the given values:
\( k = \frac{5 \times 10^{-6}}{5 \times 10^{-6} - 4 \times 10^{-6}} \)
\( k = \frac{5 \times 10^{-6}}{1 \times 10^{-6}} = 5 \)
Step 3: Validation
The calculated dielectric constant \( k \) is \( 5 \). Checking against the provided range (5,5), we confirm that the computed value fits perfectly within this range.
Thus, the dielectric constant of the slab is 5.
Approach Solution -2
Step 1: Understand the effect of a dielectric on a capacitor.
When a dielectric material is inserted between the plates of a charged capacitor, it becomes polarized, and an induced charge appears on its surfaces. This induced charge creates an electric field that opposes the original electric field due to the charges on the capacitor plates. The net electric field inside the dielectric is reduced, and consequently, the potential difference across the plates decreases, while the charge on the plates remains the same (if the capacitor is isolated).
Step 2: Relate the induced charge to the free charge and the dielectric constant.
Let \( Q \) be the free charge on the capacitor plates, and \( Q_i \) be the magnitude of the induced charge on each face of the dielectric slab. The relationship between these charges and the dielectric constant \( K \) of the material is given by: \[ Q_i = Q \left(1 - \frac{1}{K}\right) \]
Step 3: Substitute the given values into the formula.
We are given: Free charge on the capacitor plates, \( Q = 5 \times 10^{-6} \, C \) Induced charge on one face of the dielectric slab, \( Q_i = 4 \times 10^{-6} \, C \) Substituting these values into the formula: \[ 4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right) \]
Step 4: Solve for the dielectric constant \( K \).
Divide both sides by \( 5 \times 10^{-6} \): \[ \frac{4 \times 10^{-6}}{5 \times 10^{-6}} = 1 - \frac{1}{K} \] \[ \frac{4}{5} = 1 - \frac{1}{K} \] Rearrange the equation to solve for \( \frac{1}{K} \): \[ \frac{1}{K} = 1 - \frac{4}{5} \] \[ \frac{1}{K} = \frac{5}{5} - \frac{4}{5} \] \[ \frac{1}{K} = \frac{1}{5} \] Now, solve for \( K \): \[ K = 5 \] The dielectric constant of the slab is 5.
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