A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm with a die

Q: Why did we ignore the 5 mm distance given in the problem?

The relationship between the electric field and the surface charge density is independent of the plate separation. The distance was provided as a distractor to test your conceptual clarity.

Q: What is the value of $\varepsilon_0$ and do I need to memorize it?

$\varepsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \text{ F/m}$. Yes, it is a fundamental constant that you should memorize for JEE Physics.

Q: Why is the electric field $E = \sigma / (K\varepsilon_0)$ and not just $\sigma / \varepsilon_0$?

The presence of a dielectric material polarizes it, creating an opposing internal electric field. This reduces the net electric field between the plates by a factor of $K$, which is the dielectric constant.

Q: How do we know to round $5.84 \times 10^{-7}$ to $6 \times 10^{-7}$?

The question specifically asks for the value that the charge density will be 'close to'. Among the given options, $6 \times 10^{-7}$ is the nearest order of magnitude and coefficient.

Q: Does the shape of the plates (circular) matter here?

No, as long as the plates are large compared to the separation distance, the edge effects can be ignored, and the standard formula for an infinite parallel plate capacitor applies perfectly.

Question

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 	imes 10^4$ V/m, the charge density of the positive plate will be close to

Options:

(A) $6 	imes 10^{-7} 	ext{ C/m}^2$

(B) $3 	imes 10^{-7} 	ext{ C/m}^2$

(C) $3 	imes 10^4 	ext{ C/m}^2$

(D) $6 	imes 10^4 	ext{ C/m}^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a parallel plate capacitor, a fundamental component in many electronic circuits. In this problem, we are given a capacitor with circular plates separated by a distance of $5 	ext{ mm}$. Between these plates, a dielectric material with a dielectric constant $K = 2.2$ is inserted. We are also told that the electric field $E$ inside this dielectric is $3 	imes 10^4 	ext{ V/m}$. Our goal is to find the surface charge density $\sigma$ on the positive plate.

Before we dive into the math, let's address the elephant in the room: the $5 	ext{ mm}$ distance. Is it actually necessary? If we express the surface charge density as $\sigma = \frac{Q}{A}$, and substitute the formulas for capacitance $C = \frac{K\varepsilon_0 A}{d}$ and voltage $V = Ed$, we get:

$$\sigma = \frac{CV}{A} = \frac{\left(\frac{K\varepsilon_0 A}{d}ight) (Ed)}{A} = K\varepsilon_0 E$$

As you can see, both the area $A$ and the distance $d$ cancel out completely! The $5 	ext{ mm}$ distance is a classic distractor designed to test your conceptual clarity.

### The Master Equation

The relationship between the electric field inside a dielectric, the surface charge density on the plates, and the dielectric constant is beautifully simple. The presence of the dielectric polarizes the medium, creating an internal electric field that opposes the external one. This reduces the net electric field by a factor of $K$. The formula is:

$$E = \frac{\sigma}{K\varepsilon_0}$$

Since we need to find the charge density $\sigma$, we can rearrange this equation:

$$\sigma = K\varepsilon_0 E$$

### Final Calculation

Now, it's just a matter of plugging in the numbers. We know $K = 2.2$, $E = 3 	imes 10^4 	ext{ V/m}$, and the permittivity of free space $\varepsilon_0 \approx 8.85 	imes 10^{-12} 	ext{ F/m}$.

$$\sigma = 2.2 	imes (8.85 	imes 10^{-12}) 	imes (3 	imes 10^4)$$

Let's group the numbers and the powers of 10 to make the calculation easier:

$$\sigma = (2.2 	imes 3) 	imes 8.85 	imes 10^{-12} 	imes 10^4$$

$$\sigma = 6.6 	imes 8.85 	imes 10^{-8}$$

Multiplying $6.6$ by $8.85$ gives us exactly $58.41$. So, we have:

$$\sigma = 58.41 	imes 10^{-8} 	ext{ C/m}^2$$

To match the format of the options, we can rewrite this in standard scientific notation:

$$\sigma = 5.841 	imes 10^{-7} 	ext{ C/m}^2$$

Looking at the given options, the closest value is **$6 	imes 10^{-7} 	ext{ C/m}^2$**. This is a great example of how JEE problems often require you to find the nearest approximate answer rather than an exact match.

Select Answer:

Visualized Solution

\text{Visualizing the Capacitor}

\text{Electric Field in Dielectric}

\text{Is distance } d \text{ needed?}

\text{Rearranging for } \sigma

\text{Substituting Values}

\text{Calculating } \sigma

\text{Final Result}

\text{Food for Thought}

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Master Equation

Final Calculation

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