Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm2 are joined in such a way that <img alt="\alpha \neq" src="https://entrancecorner.oncodecogs.com/gif.lat

As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm2 are joined in such a way that \alpha \neq b.
The equivalent capacitance of the combination is x \epsilon_0  F. The value of x is _______.

                      

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

As per the arrangement given, distance between the capacitor plates are a and b and a \neq  b
using the diagram we can write
b = 5 – a – 1 = (4 – a) in mm

as we know capacitance of capacitor C  =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} 

and in series arrangement

\begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} \\ & \frac{1}{\mathrm{C}_{\text {eq }}}=\frac{\mathrm{a}}{\varepsilon_0 \mathrm{~A}}+\frac{4-\mathrm{a}}{\varepsilon_0 \mathrm{~A}}=\frac{4(\text { in mm })}{\varepsilon_0 \mathrm{~A}} \\ & \text { or } \mathrm{C}_{\text {eq }}=\frac{\varepsilon_0 \mathrm{~A}}{4(\mathrm{~mm})} \end{aligned}

Given \mathrm{A}=200 \mathrm{~cm}^2\\ C_{e q}=\frac{\varepsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}}

=\varepsilon_0 50 \times 10^{-1} or \mathrm{C}_{\text {eq }}=5 \varepsilon_0 farad

Therefore n=5 

 

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question