A parallel plate capacitor was made with two rectangular plates, each with a length of $l=3 \mathrm{~cm}$…
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 only
- C only
- B and D only
- $C$ 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
Asked in: JEE Main 2025 (24 Jan Shift 1)