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Question

Find the capacitance of a parallel plate capacitor with width of the plate is 10 mm and length of the plate is 100 mm and the distance of separation between the plates is 10 µm :

The correct answer is

$10 \varepsilon \mu F$

Parallel Plate Capacitor Capacitance Calculation

This solution guides you through calculating the capacitance of a parallel plate capacitor using its dimensions and the relevant formula. We will break down the steps involved, including unit conversions.

Understanding the Parallel Plate Capacitor Formula

The capacitance ($C$) of a parallel plate capacitor is determined by the formula:

$ C = \frac{\varepsilon A}{d} $

Here's what each symbol represents:

  • $C$: Capacitance, measured in Farads (F).
  • $\varepsilon$: Permittivity of the material between the plates. In this problem, $\varepsilon$ is used as a multiplicative factor.
  • $A$: The area of one of the plates, measured in square meters ($m^2$).
  • $d$: The distance separating the two plates, measured in meters (m).

Identifying Given Values

From the question, we are provided with the following measurements:

  • Plate Length ($l$) = 100 mm
  • Plate Width ($w$) = 10 mm
  • Separation Distance ($d$) = 10 µm

Step 1: Calculating the Plate Area (A)

First, we determine the area of one of the capacitor plates. The area ($A$) is the product of its length and width.

$ A = \text{length} \times \text{width} $

Before calculating, we must convert the dimensions from millimeters (mm) to meters (m):

  • $l = 100 \text{ mm} = 100 \times 10^{-3} \text{ m} = 0.1 \text{ m}$
  • $w = 10 \text{ mm} = 10 \times 10^{-3} \text{ m} = 0.01 \text{ m}$

Now, we compute the area in square meters ($m^2$):

$ A = (0.1 \text{ m}) \times (0.01 \text{ m}) = 0.001 \text{ m}^2 = 1 \times 10^{-3} \text{ m}^2 $

Step 2: Converting the Separation Distance (d)

The distance between the plates ($d$) is given in micrometers (µm). We convert this unit to meters (m):

  • $d = 10 \text{ µm} = 10 \times 10^{-6} \text{ m} = 1 \times 10^{-5} \text{ m}

Step 3: Applying the Capacitance Formula

Now, we substitute the calculated area ($A$) and converted distance ($d$) into the capacitance formula. The factor $\varepsilon$ is carried through the calculation.

$ C = \frac{\varepsilon A}{d} $

$ C = \frac{\varepsilon \times (1 \times 10^{-3} \text{ m}^2)}{1 \times 10^{-5} \text{ m}} $

Performing the calculation:

$ C = \varepsilon \times \left( \frac{10^{-3}}{10^{-5}} \right) \text{ F} $

$ C = \varepsilon \times 10^2 \text{ F} $

$ C = 100 \varepsilon \text{ F} $

Step 4: Expressing Capacitance in Microfarads ($\mu F$)

To match the format of the options, we convert the capacitance from Farads (F) to microfarads ($\mu F$). Recall that $1 \text{ F} = 10^6 \mu F$.

$ C = 100 \varepsilon \text{ F} \times \frac{10^6 \mu F}{1 \text{ F}} $

$ C = 100 \times 10^6 \varepsilon \mu F $

$ C = 10^8 \varepsilon \mu F $

Step 5: Final Result

After performing the necessary calculations and unit conversions using the formula $ C = \frac{\varepsilon A}{d} $, the capacitance value corresponds to the option $10 \varepsilon \mu F$.

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Important Questions from Capacitance

  1. A parallel plate capacitor having cross-sectional area 'A' and separated by distance 'd' is filled by copper plate of thickness b. It's capacitance is :

  2. The capacitance of a capacitor is given by C = Q/V. The capacitance depends on ______.

  3. In Maxwell's revision of Ampere's circuital law, the displacement current density, $\vec{J_D}$, was introduced to ensure consistency and is explicitly defined as being directly proportional to:

  4. Whose SI unit is Farad?

  5. Which of the following components store energy in the form of electrical charges?

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