4274 Views

If the distance between plates of a capacitor having capacity C and charge Q is doubled then the work done will be : (a)Q^2/4C (b)Q^2/2C


Kusum yadav 11th Jan, 2019
Answer (1)
Akhila Nagasree 11th Jan, 2019

Hello,

The energy initially stored by the capacitor is (1/2 C*V^2) or( Q^2 / 2*C)

When the plate separation is doubled, the charge remains constant and the capacitance is halved. The work done in separating the plates is the difference in energy stored:this is

Q^2 / 2*0.5*C - Q^2 / 2*CT

This represents double the initial energy less the initial energy. The work done in adding this energy is Q^2 / 2*C.that is answer is option B

Hope this helps......

1 Comment
Comments (1)
11th Jan, 2019
Yeah...i got it...thank u 
Reply

Related Questions

UPES Integrated LLB Admission...
Apply
Ranked #28 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS University Rankings | 16.6 LPA Highest CTC
Jindal Global Law School Admi...
Apply
Ranked #1 Law School in India & South Asia by QS- World University Rankings | Merit cum means scholarships | Application Deadline: 30th Nov'24
Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities In QS Asia Rankings 2025 | Scholarships worth 210 CR
Great Lakes PGPM & PGDM 2025
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 1st Dec 2024
ICFAI Business School-IBSAT 2024
Apply
9 IBS Campuses | Scholarships Worth Rs 10 CR
Nirma University Law Admissio...
Apply
Grade 'A+' accredited by NAAC
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books