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EEJ MAIN Mathematics QUESTION #10985
Question 1

A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure with a dielectric constant $K=5$, the percentage increase in the capacitance is ________.

  • 66.67

    ✔️

  • 33.34

  • 400

  • 200

Correct Answer Explanation

Step 1: Initial Capacitance

The initial capacitance of the parallel plate air capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

Step 2: Equivalent Capacitance with Dielectric

When a dielectric of constant $K = 5$ fills half the distance ($\frac{d}{2}$), the system behaves as two capacitors connected in series:

  • Air-filled portion ($C_1$):

    $$C_1 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2\varepsilon_0 A}{d} = 2C$$

  • Dielectric-filled portion ($C_2$):

    $$C_2 = \frac{K \varepsilon_0 A}{\frac{d}{2}} = \frac{5 \cdot 2\varepsilon_0 A}{d} = 10C$$

Using the equivalent series capacitance formula:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$$
$$\frac{1}{C_{\text{eq}}} = \frac{1}{2C} + \frac{1}{10C} = \frac{5 + 1}{10C} = \frac{6}{10C} = \frac{3}{5C}$$
$$C_{\text{eq}} = \frac{5}{3}C$$

Step 3: Percentage Increase

$$\text{Percentage Increase} = \left(\frac{C_{\text{eq}} - C}{C}\right) \times 100\%$$
$$\text{Percentage Increase} = \left(\frac{\frac{5}{3}C - C}{C}\right) \times 100\% = \frac{2}{3} \times 100\% \approx 66.67\%$$