A parallel plate capacitor was made with two rectangular plates, each with a length of $l=3 \mathrm{~cm}$ and breath of $\mathrm{b}=1 \mathrm{~cm}$. The distance between the plates is $3 \mu \mathrm{~m}$. Out of the following, which are the ways to increase the capacitance by a factor of 10 ?

A. $l=30 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}$

B. $l=3 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=30 \mu \mathrm{~m}$

C. $l=6 \mathrm{~cm}, \mathrm{~b}=5 \mathrm{~cm}, \mathrm{~d}=3 \mu \mathrm{~m}$

D. $l=1 \mathrm{~cm}, \mathrm{~b}=1 \mathrm{~cm}, \mathrm{~d}=10 \mu \mathrm{~m}$

E. $l=5 \mathrm{~cm}, \mathrm{~b}=2 \mathrm{~cm}, \mathrm{~d}=1 \mu \mathrm{~m}$

Choose the correct answer from the options given below:

<p><strong>The capacitance of a parallel plate capacitor is given by</strong></p> <p>$C = \frac{\epsilon_0 A}{d},$</p> <p><strong>where:</strong> </p> <p><p>$\epsilon_0$ is the permittivity of free space, </p></p> <p><p>$A$ is the area of a plate, and </p></p> <p><p>$d$ is the separation between the plates.</p></p> <p>A change in the capacitor’s dimensions will change its capacitance by a factor of</p> <p>$$ \frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}. $$</p> <p>The initial dimensions are: </p> <p><p>Length: $l = 3\,\text{cm} = 0.03\,\text{m},$ </p></p> <p><p>Breadth: $b = 1\,\text{cm} = 0.01\,\text{m},$ </p></p> <p><p>Hence, the original area is </p></p> <p>$A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2,$ </p> <p>Plate separation: $d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.$</p> <p><strong>For each modification, we compare the new factor:</strong></p> <p><strong>Case C:</strong> </p> <p><p>New dimensions: </p> <p>$l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.$ </p></p> <p><p>New area: </p> <p>$A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2.$</p></p> <p><p>Ratio of areas: </p> <p>$$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10. $$</p></p> <p><p>The plate separation is unchanged: </p> <p>$$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1. $$</p></p> <p><p>Overall factor: </p> <p>$10 \times 1 = 10.$</p></p> <p><strong>Case E:</strong> </p> <p><p>New dimensions: </p> <p>$l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.$ </p></p> <p><p>New area: </p> <p>$A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2.$</p></p> <p><p>Ratio of areas: </p> <p>$$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33. $$</p></p> <p><p>New plate separation: </p> <p>$d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},$ </p> <p>so the ratio of separations is </p> <p>$$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3. $$</p></p> <p><p>Overall factor: </p> <p>$3.33 \times 3 \approx 10.$</p></p> <p><strong>Both Case C and Case E yield a capacitance that is 10 times the original value.</strong></p> <p><strong>Thus, the correct modifications are those in Case C and Case E.</strong></p>