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:

• A

  A only
• B

  C only
• C

  B and D only
• Correct answer
  D

  $C$ and $E$ only

Solution

We know $C=\frac{A \varepsilon_0}{d}$ $=\frac{b \ell \varepsilon_0}{d}$ So to increase the capacitance by 10 factor $\left(\frac{A}{d}\right)$ has to increase by 10 fator. For option (A) $\quad C^{\prime}=\frac{(30 \ell) b \varepsilon_0}{\left(\frac{d}{3}\right)}=30 \mathrm{C}$ For option (B) $\quad C^{\prime}=\frac{\ell b \varepsilon_0}{10 d}=\frac{C}{10}$ For option (C) $\quad C^{\prime}=\frac{(2 \ell) 5 b \varepsilon_0}{d}=10 \mathrm{C}$ For option (D) $\quad C^{\prime}=\frac{\left(\frac{\ell}{3}\right) b \varepsilon_0}{\left(\frac{10 d}{3}\right)}=\frac{C}{10}$ For option (E) $\quad C^{\prime}=\frac{\left(\frac{\ell}{3}\right) 5(2 b) \varepsilon_0}{\left(\frac{d}{3}\right)}=10 C$ Clearly (C) and (E) are the situation for 10 C

Hence, the correct answer is option: 'D'.