Capacitor with Dielectric — Quick Trick to Find K (JEE Main Concept) ⚡

 

❓ Question

A parallel plate capacitor has a charge of $5 \times 10^{-6}\ \text{C}$.
A dielectric slab is inserted between the plates and almost fills the space between them.
If the induced charge on one face of the slab is $4 \times 10^{-6}\ \text{C}$, then find the dielectric constant ($K$) of the slab.

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Recall the concept

When a dielectric slab is inserted into a charged capacitor:

• The free charge $Q$ on the plates remains the same (if disconnected from battery).

• The dielectric becomes polarized, producing induced charge $Q'$ on its surfaces.

• Relation between induced charge and free charge:

$$Q^{\mathrm{\prime }}=Q(1-\frac{1}{K})$$

where $K$ = dielectric constant.

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Step 2 — Substitute given values

Given:

$$Q=5\times {10}^{-6}\text{ C},Q^{\mathrm{\prime }}=4\times {10}^{-6}\text{ C}\mathrm{.}$$

So,

$$4\times {10}^{-6}=5\times {10}^{-6}(1-\frac{1}{K})$$

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Step 3 — Simplify the equation

Cancel $10^{-6}$ from both sides:

$$4 = 5\left(1 - \frac{1}{K}\right)$$
$$\frac{4}{5} = 1 - \frac{1}{K}$$ $$\frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}$$ $$K = 5$$

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🧮 Image Solution

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✅ Conclusion & Concept Recap

✅ Final Answer:

$$\boxed{K = 5}$$

Concept Recap:

• The dielectric slab reduces the electric field inside by a factor of $K$.

• Induced charge $Q'$ always acts to oppose the field from free charge.

• Larger $K$ → stronger polarization → more induced charge.

• Formula to remember:

  $$Q^{\mathrm{\prime }}=Q(1-\frac{1}{K})$$

  or equivalently,

  $$K=\frac{1}{1-Q^{\mathrm{\prime }}\mathrm{/}\mathrm{Q}}$$