A capacitor of 10 μ F capacitance whose plates are separated by 10 ~mm through air and each plate has area 4 ~cm^2 is now filled equally with two dielectric media of K_1=2, ~K_2=3 respectively as shown in figure. If new

Correct answer: 0 aligned & C_ eq =C_1+C_2 & C_1= 2 _0 ~A 2 × d =10 μ F & C_2= 3 _0 ~A 2 ~d =15 μ F & C_ eq =25 μ F aligned Now the charge on aligned & C_1=10 ~V μ c & C_2=1.5 ~V μ C . aligned Now force between the plates [F=(Q^2)/(2 A ∈_0)] aligned & 100 ~V^2 × 10^ -12 2 × 2 × 10^ -4 _0 + 225 ~V^2 × 10^ -12 2 × 2 × 10^ -4 × _0 =8 & 325 ~V^2=8 × 4 × 10^ -4 × 8.85 & ~V^2= 32 × 8.85 × 10^ -4 325 & V= 283.2 × 10^ -4 325 & ~V=0.93 × 10^ -2 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Physics
Chapter: Capacitance

JEE Main 2024 · 6 April · Evening Shift · Physics · Capacitance

A capacitor of 10 μ F capacitance whose plates are separated by 10...

A capacitor of

10 \mu \mathrm{F}

capacitance whose plates are separated by

10 \mathrm{~mm}

through air and each plate has area

4 \mathrm{~cm}^2

is now filled equally with two dielectric media of

\mathrm{K}_1=2, \mathrm{~K}_2=3

respectively as shown in figure. If new force between the plates is

8 \mathrm{~N}

. The supply voltage is _________

\times 10^{-4} \mathrm{~V}

A capacitor of 10 μ F capacitance whose plates are separated by 10 ~mm through air and each plate has area 4 ~cm^2 is

(we modified language of question to make it correct)

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \\ & \mathrm{C}_1=\frac{2 \epsilon_0 \mathrm{~A}}{2 \times \mathrm{d}}=10 \mu \mathrm{F} \\ & \mathrm{C}_2=\frac{3 \epsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=15 \mu \mathrm{F} \\ & \mathrm{C}_{\mathrm{eq}}=25 \mu \mathrm{F} \end{aligned}

Now the charge on

\begin{aligned} & \mathrm{C}_1=10 \mathrm{~V} \mu \mathrm{c} \\ & \mathrm{C}_2=1.5 \mathrm{~V} \mu \mathrm{C} . \end{aligned}

Now force between the plates \left[F=\frac{Q^2}{2 A \in_0}\right]

\begin{aligned} & \frac{100 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \epsilon_0}+\frac{225 \mathrm{~V}^2 \times 10^{-12}}{2 \times 2 \times 10^{-4} \times \epsilon_0}=8 \\ & 325 \mathrm{~V}^2=8 \times 4 \times 10^{-4} \times 8.85 \\ & \mathrm{~V}^2=\frac{32 \times 8.85 \times 10^{-4}}{325} \\ & \therefore \mathrm{V}=\sqrt{\frac{283.2 \times 10^{-4}}{325}} \\ & \mathrm{~V}=0.93 \times 10^{-2} \end{aligned}

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