Question
Download Solution PDFIf the distance between the plates of a parallel plate capacitor is increased 10 times and the area is reduced to one-fourth, then its capacitance _______.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The capacitance of a parallel plate capacitor is given by:
\(C={ϵ_oA\over d}\)
where, ϵo = Permittivity of free space
A = Area of parallel plate capacitor
d = Distance between parallel plate capacitor
Calculation
Given, \(d_2=10d_1\)
\(A_2={A_1\over 4}\)
\({C_2\over C_1}={A_2\over d_2}\times {d_1\over A_1}\)
\({C_2\over C_1}={A_1\over 4\times 10\times d_1}\times {d_1\over A_1}\)
\(C_2={C_1\over 40}\)
Thus, the new capacitance becomes \(\frac{1}{40}\) times of initial capacitance.
Last updated on Jun 16, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.