Find Dielectric Constant Using Induced Charge Method

 

❓ 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

✍️ Short Explanation

This problem is based on:

👉 Dielectrics
👉 Polarization
👉 Capacitors with dielectric slab.

Main idea:

Induced charge on dielectric surface:

$$\boxed{ q' = q\left(1-\frac1K\right) }$$

where:

• $q$ = free charge on capacitor
• $q'$= induced bound charge
• $K$ = dielectric constant

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🔷 Step 1 — Write Formula 💯

Given:

$$q=5\times10^{-6}\text{ C}$$

Induced charge:

$$q'=4\times10^{-6}\text{ C}$$

Use:

$$q' = q\left(1-\frac1K\right)$$

Substitute values:

$$4\times10^{-6} = 5\times10^{-6} \left(1-\frac1K\right)$$

Cancel:

$$4 = 5\left(1-\frac1K\right)$$

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🔷 Step 2 — Solve for $K$

$$\frac45 = 1-\frac1K$$
$$\frac1K = 1-\frac45$$
$$\frac1K = \frac15$$

Thus:

$$\boxed{ K=5 }$$

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🔷 Step 3 — Physical Meaning

Larger dielectric constant:

✔ More polarization

✔ Larger induced charge

✔ Greater capacitance

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Induced charge ko directly equal to free charge maan lena

❌ Formula me $1-\frac1K$ ki jagah $\frac1K$ use kar dena

Remember:

$$\boxed{ q_{\text{induced}} = q\left(1-\frac1K\right) }$$

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✅ Final Answer

$$\boxed{ 5 }$$

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