A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 ~cm and breath of b=1 ~cm . The distance between the plates is 3 μ ~m . Out of the following, which are the ways to increase the

Correct answer: D. C and E only We know C=(A _0)/(d) =(b _0)/(d) So to increase the capacitance by 10 factor ((A)/(d)) has to increase by 10 fator. For option (A) C^ =((30 ) b _0)/(((d)/(3)))=30 C For option (B) C^ =( b _0)/(10 d)=(C)/(10) For option (C) C^ =((2 ) 5 b _0)/(d)=10 C For option (D) C^ = (( )/(3)) b _0 ((10 d)/(3)) =(C)/(10) For option (E) C^ =((( )/(3)) 5(2 b) _0)/(((d)/(3)))=10 C Clearly (C) and (E) are the situation for 10 C

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Physics
Chapter: Capacitance

JEE Main 2025 · 24 January · Morning Shift · Physics · Capacitance

A parallel plate capacitor was made with two rectangular plates,...

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}

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

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

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

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 parallel plate capacitor was made with two rectangular plates, each with a length of l=3 ~cm and breath of b=1 ~cm .

A parallel plate capacitor was made with two rectangular plates, each with a length of l=3 ~cm and breath of b=1 ~cm .

Options

AA only
BC only
CB and D only
D $C$ and $E$ only

Correct answer

D. $C$ and $E$ only

Step-by-step 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

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